💡Okay, so I have this problem here: I need to find all real numbers ( k ) such that the norm of ( k ) times the vector ( begin{pmatrix} 3 -4 end{pmatrix} ) minus the vector ( begin{pmatrix} 5 8 end{pmatrix} ) equals ( 3sqrt{13} ). Hmm, that sounds a bit complicated, but let me break it down step by step.First, I remember that the norm of a vector ( begin{pmatrix} a b end{pmatrix} ) is calculated as ( sqrt{a^2 + b^2} ). So, I need to compute the norm of the vector ( k begin{pmatrix} 3 -4 end{pmatrix} - begin{pmatrix} 5 8 end{pmatrix} ). Let me write that out.Multiplying the vector by ( k ) gives me ( begin{pmatrix} 3k -4k end{pmatrix} ). Then, subtracting ( begin{pmatrix} 5 8 end{pmatrix} ) from that, I get ( begin{pmatrix} 3k - 5 -4k - 8 end{pmatrix} ). So, the vector inside the norm is ( begin{pmatrix} 3k - 5 -4k - 8 end{pmatrix} ).Now, I need to find the norm of this vector. Using the formula for the norm, it should be ( sqrt{(3k - 5)^2 + (-4k - 8)^2} ). Let me compute each part separately.First, ( (3k - 5)^2 ). Expanding that, I get ( 9k^2 - 30k + 25 ). Okay, that seems right. Then, ( (-4k - 8)^2 ). Expanding that, I get ( 16k^2 + 64k + 64 ). Wait, let me double-check that. Squaring ( -4k - 8 ) gives ( (-4k)^2 + 2*(-4k)*(-8) + (-8)^2 ), which is ( 16k^2 + 64k + 64 ). Yeah, that's correct.So, adding these two results together: ( 9k^2 - 30k + 25 + 16k^2 + 64k + 64 ). Combining like terms, ( 9k^2 + 16k^2 = 25k^2 ), ( -30k + 64k = 34k ), and ( 25 + 64 = 89 ). So, the expression under the square root simplifies to ( 25k^2 + 34k + 89 ).The equation we have is ( sqrt{25k^2 + 34k + 89} = 3sqrt{13} ). To get rid of the square root, I can square both sides of the equation. That gives me ( 25k^2 + 34k + 89 = (3sqrt{13})^2 ). Calculating the right side, ( (3sqrt{13})^2 = 9 * 13 = 117 ).So now, the equation is ( 25k^2 + 34k + 89 = 117 ). Subtracting 117 from both sides to set the equation to zero: ( 25k^2 + 34k + 89 - 117 = 0 ). Simplifying that, ( 25k^2 + 34k - 28 = 0 ). Wait, hold on, 89 - 117 is -28, right? Let me check: 117 - 89 is 28, so yeah, it's -28. So, the quadratic equation is ( 25k^2 + 34k - 28 = 0 ).Now, I need to solve this quadratic equation for ( k ). I can use the quadratic formula, which is ( k = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 25 ), ( b = 34 ), and ( c = -28 ).Calculating the discriminant first: ( b^2 - 4ac = 34^2 - 4*25*(-28) ). Let's compute each part. ( 34^2 = 1156 ). Then, ( 4*25 = 100 ), and ( 100*(-28) = -2800 ). So, the discriminant becomes ( 1156 - (-2800) = 1156 + 2800 = 3956 ).Wait, that seems a bit high. Let me double-check my calculations. ( 34^2 ) is indeed 1156. ( 4*25 = 100 ), correct. ( 100*(-28) = -2800 ), right. So, subtracting a negative is adding, so 1156 + 2800 is 3956. Hmm, okay, that's correct.So, the discriminant is 3956. Now, taking the square root of 3956. Hmm, let me see if 3956 is a perfect square. Let me try dividing by some squares. 3956 divided by 4 is 989, which is not a perfect square. 989 divided by 13 is 76.07, which isn't an integer. Maybe 3956 is 63^2? 63^2 is 3969, which is just a bit higher. So, 63^2 is 3969, so 3956 is 63^2 minus 13, which isn't a perfect square. So, the square root of 3956 is irrational. Let me compute it approximately. 63^2 is 3969, so sqrt(3956) is approximately 63 - (13)/(2*63) = 63 - 13/126 ≈ 63 - 0.103 ≈ 62.897.But since we're dealing with exact values, I can leave it as sqrt(3956). Wait, but maybe I made a mistake earlier because the discriminant seems quite large, which might lead to a complicated solution. Let me double-check the earlier steps to make sure I didn't make an error.Starting from the beginning: the vector is ( k begin{pmatrix} 3 -4 end{pmatrix} - begin{pmatrix} 5 8 end{pmatrix} ), which is ( begin{pmatrix} 3k - 5 -4k - 8 end{pmatrix} ). The norm squared is ( (3k - 5)^2 + (-4k - 8)^2 ). Expanding those, I get 9k² - 30k + 25 + 16k² + 64k + 64. Adding them up: 25k² + 34k + 89. So, that's correct.Setting that equal to ( (3sqrt{13})^2 = 9*13 = 117 ). So, 25k² + 34k + 89 = 117. Subtracting 117: 25k² + 34k - 28 = 0. So, that's correct.So, the quadratic is correct, and the discriminant is 34² - 4*25*(-28) = 1156 + 2800 = 3956. So, sqrt(3956). Hmm, maybe I can factor 3956 to see if it simplifies. Let's try dividing by 4: 3956 ÷ 4 = 989. 989 ÷ 13 = 76.07, which isn't an integer. 989 ÷ 7 = 141.28, not integer. 989 ÷ 11 = 89.909, nope. 989 ÷ 17 = 58.176, nope. Maybe 989 is prime? Let me check: 989 ÷ 23 = 43. So, 23*43 = 989. So, 3956 = 4*23*43. So, sqrt(3956) = 2*sqrt(23*43) = 2*sqrt(989). So, it doesn't simplify further.So, plugging back into the quadratic formula: ( k = frac{-34 pm 2sqrt{989}}{50} ). Simplifying that, I can factor out a 2 from numerator and denominator: ( k = frac{-17 pm sqrt{989}}{25} ).Wait, but hold on a second. The problem asks for real numbers ( k ). Since the discriminant is positive (3956 is positive), there are two real solutions. So, the solutions are ( k = frac{-17 + sqrt{989}}{25} ) and ( k = frac{-17 - sqrt{989}}{25} ).But wait, earlier in my initial calculation, I thought the discriminant was negative, but that was because I made a mistake in subtracting 117. Let me go back to that step. Wait, in my first attempt, I had 25k² + 34k + 89 = 117, which led to 25k² + 34k - 28 = 0. But in my initial thought process, I thought it was 25k² + 34k + 50 = 0, which was incorrect because I subtracted 117 incorrectly. So, that was my mistake. The correct equation is 25k² + 34k - 28 = 0, which has a positive discriminant, so two real solutions.Therefore, the real numbers ( k ) that satisfy the equation are ( k = frac{-17 + sqrt{989}}{25} ) and ( k = frac{-17 - sqrt{989}}{25} ).Wait, but let me double-check the discriminant calculation again because 34² is 1156, and 4ac is 4*25*(-28) = -2800. So, discriminant is 1156 - (-2800) = 1156 + 2800 = 3956. Yes, that's correct.So, the solutions are ( k = frac{-34 pm sqrt{3956}}{50} ). Simplifying sqrt(3956) as 2*sqrt(989), so ( k = frac{-34 pm 2sqrt{989}}{50} ). Dividing numerator and denominator by 2: ( k = frac{-17 pm sqrt{989}}{25} ).So, those are the two real solutions. Therefore, the real numbers ( k ) that satisfy the given equation are ( frac{-17 + sqrt{989}}{25} ) and ( frac{-17 - sqrt{989}}{25} ).Wait, but let me verify if these solutions actually satisfy the original equation. Let me pick one of them and plug it back in. Let's take ( k = frac{-17 + sqrt{989}}{25} ).First, compute ( 3k - 5 ): ( 3*frac{-17 + sqrt{989}}{25} - 5 = frac{-51 + 3sqrt{989}}{25} - frac{125}{25} = frac{-51 - 125 + 3sqrt{989}}{25} = frac{-176 + 3sqrt{989}}{25} ).Then, compute ( -4k - 8 ): ( -4*frac{-17 + sqrt{989}}{25} - 8 = frac{68 - 4sqrt{989}}{25} - frac{200}{25} = frac{68 - 200 - 4sqrt{989}}{25} = frac{-132 - 4sqrt{989}}{25} ).Now, the norm squared is ( (frac{-176 + 3sqrt{989}}{25})^2 + (frac{-132 - 4sqrt{989}}{25})^2 ). Let me compute each term.First term: ( (frac{-176 + 3sqrt{989}}{25})^2 = frac{(-176)^2 + (3sqrt{989})^2 - 2*176*3sqrt{989}}{625} = frac{30976 + 9*989 - 1056sqrt{989}}{625} ).Calculating 9*989: 9*900=8100, 9*89=801, so total 8100+801=8901. So, first term becomes ( frac{30976 + 8901 - 1056sqrt{989}}{625} = frac{39877 - 1056sqrt{989}}{625} ).Second term: ( (frac{-132 - 4sqrt{989}}{25})^2 = frac{(-132)^2 + (4sqrt{989})^2 + 2*132*4sqrt{989}}{625} = frac{17424 + 16*989 + 1056sqrt{989}}{625} ).Calculating 16*989: 10*989=9890, 6*989=5934, so total 9890+5934=15824. So, second term becomes ( frac{17424 + 15824 + 1056sqrt{989}}{625} = frac{33248 + 1056sqrt{989}}{625} ).Adding both terms: ( frac{39877 - 1056sqrt{989} + 33248 + 1056sqrt{989}}{625} = frac{39877 + 33248}{625} = frac{73125}{625} ).Simplifying ( frac{73125}{625} ): 625*117 = 73125. So, the norm squared is 117, which is equal to ( (3sqrt{13})^2 ). Therefore, the norm is indeed ( 3sqrt{13} ). So, the solution ( k = frac{-17 + sqrt{989}}{25} ) is correct.Similarly, the other solution ( k = frac{-17 - sqrt{989}}{25} ) should also satisfy the equation. Let me briefly check that.Compute ( 3k - 5 ): ( 3*frac{-17 - sqrt{989}}{25} - 5 = frac{-51 - 3sqrt{989}}{25} - frac{125}{25} = frac{-51 - 125 - 3sqrt{989}}{25} = frac{-176 - 3sqrt{989}}{25} ).Compute ( -4k - 8 ): ( -4*frac{-17 - sqrt{989}}{25} - 8 = frac{68 + 4sqrt{989}}{25} - frac{200}{25} = frac{68 - 200 + 4sqrt{989}}{25} = frac{-132 + 4sqrt{989}}{25} ).Now, the norm squared is ( (frac{-176 - 3sqrt{989}}{25})^2 + (frac{-132 + 4sqrt{989}}{25})^2 ).First term: ( (frac{-176 - 3sqrt{989}}{25})^2 = frac{(-176)^2 + (3sqrt{989})^2 + 2*176*3sqrt{989}}{625} = frac{30976 + 8901 + 1056sqrt{989}}{625} = frac{39877 + 1056sqrt{989}}{625} ).Second term: ( (frac{-132 + 4sqrt{989}}{25})^2 = frac{(-132)^2 + (4sqrt{989})^2 - 2*132*4sqrt{989}}{625} = frac{17424 + 15824 - 1056sqrt{989}}{625} = frac{33248 - 1056sqrt{989}}{625} ).Adding both terms: ( frac{39877 + 1056sqrt{989} + 33248 - 1056sqrt{989}}{625} = frac{39877 + 33248}{625} = frac{73125}{625} = 117 ). So, again, the norm squared is 117, which is correct.Therefore, both solutions satisfy the original equation. So, the real numbers ( k ) that satisfy the given condition are ( frac{-17 + sqrt{989}}{25} ) and ( frac{-17 - sqrt{989}}{25} ).I think that's it. I went through the steps carefully, checked my calculations, and verified the solutions by plugging them back into the original equation. It seems correct.

💡To determine the most accurate value for ( K ) such that every digit is significant, we start by calculating the range of possible values for ( K ) given the error. 1. **Calculate the upper and lower bounds of ( K )**: [ K_{text{upper}} = 3.68547 + 0.00256 = 3.68803 ] [ K_{text{lower}} = 3.68547 - 0.00256 = 3.68291 ]2. **Round these bounds to the nearest hundredth**: - ( K_{text{upper}} = 3.68803 ) rounds to ( 3.69 ) - ( K_{text{lower}} = 3.68291 ) rounds to ( 3.68 )Since the rounded values differ at the hundredth place, we consider the next significant digit. Rounding both bounds to the nearest tenth gives ( 3.7 ) for both, ensuring all digits are significant.Thus, the most accurate value that can be announced for ( K ) is ( 3.7 ).

💡Okay, so I have this problem about a geometric sequence. Let me try to understand what it's asking. We have a geometric sequence {a_n}, and we're given that the second term, a_2, is 1/4, and the sixth term, a_6, is 4. We need to find the product of the first seven terms, denoted as T_7. The options given are A: 1, B: 1 or -1, C: 2, D: 2 or -2.Alright, so first, let's recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as q. So, in general, a_n = a_1 * q^(n-1), where a_1 is the first term.Given that, let's write down the expressions for a_2 and a_6.a_2 = a_1 * q^(2-1) = a_1 * q = 1/4a_6 = a_1 * q^(6-1) = a_1 * q^5 = 4So, we have two equations:1) a_1 * q = 1/42) a_1 * q^5 = 4Hmm, so we can use these two equations to solve for a_1 and q. Maybe by dividing equation 2 by equation 1, we can eliminate a_1.Let's try that:(a_1 * q^5) / (a_1 * q) = 4 / (1/4)Simplify the left side: a_1 cancels out, and q^5 / q = q^4Right side: 4 divided by 1/4 is 16.So, q^4 = 16Therefore, q is the fourth root of 16. The fourth roots of 16 are 2 and -2 because 2^4 = 16 and (-2)^4 = 16 as well.So, q can be 2 or -2.Now, let's find a_1 for each case.Case 1: q = 2From equation 1: a_1 * 2 = 1/4 => a_1 = (1/4) / 2 = 1/8Case 2: q = -2From equation 1: a_1 * (-2) = 1/4 => a_1 = (1/4) / (-2) = -1/8So, we have two possibilities:Either a_1 = 1/8 and q = 2, or a_1 = -1/8 and q = -2.Now, we need to find T_7, which is the product of the first seven terms.Let me recall that the product of terms in a geometric sequence can be expressed as:T_n = (a_1 * a_n)^(n/2)Wait, is that correct? Let me think.Actually, the product of the first n terms of a geometric sequence is given by:T_n = a_1^n * q^(n(n-1)/2)Alternatively, since each term is a multiple of the previous one by q, the product would be a_1 * (a_1 q) * (a_1 q^2) * ... * (a_1 q^(n-1)).Which simplifies to a_1^n * q^(0 + 1 + 2 + ... + (n-1)).The exponent of q is the sum of the first (n-1) integers, which is n(n-1)/2.So, T_n = a_1^n * q^(n(n-1)/2)Alternatively, since a_n = a_1 q^(n-1), so a_1 = a_n / q^(n-1). Plugging that into T_n:T_n = (a_n / q^(n-1))^n * q^(n(n-1)/2) = a_n^n / q^(n(n-1)) * q^(n(n-1)/2) = a_n^n / q^(n(n-1)/2)But maybe it's simpler to stick with the first expression.So, T_7 = a_1^7 * q^(7*6/2) = a_1^7 * q^21Alternatively, since 7*6/2 is 21.So, let's compute T_7 for both cases.Case 1: a_1 = 1/8, q = 2T_7 = (1/8)^7 * 2^21Compute (1/8)^7: 1/(8^7) = 1/(2^21) because 8 is 2^3, so 8^7 = (2^3)^7 = 2^21.So, (1/8)^7 = 1/2^21Then, 2^21 is just 2^21.So, T_7 = (1/2^21) * 2^21 = 1Case 2: a_1 = -1/8, q = -2T_7 = (-1/8)^7 * (-2)^21Compute (-1/8)^7: (-1)^7 / 8^7 = -1 / 2^21Compute (-2)^21: (-2)^21 = -2^21 because the exponent is odd.So, T_7 = (-1 / 2^21) * (-2^21) = (-1) * (-1) * (2^21 / 2^21) = 1 * 1 = 1Wait, so in both cases, T_7 is 1?But wait, let me double-check.In case 2, a_1 = -1/8, q = -2.So, T_7 = (-1/8)^7 * (-2)^21(-1/8)^7 = (-1)^7 / 8^7 = -1 / 2^21(-2)^21 = (-1)^21 * 2^21 = -1 * 2^21So, multiplying them: (-1 / 2^21) * (-2^21) = (-1) * (-1) * (2^21 / 2^21) = 1 * 1 = 1Yes, so in both cases, T_7 is 1.Wait, but the options include 1, 1 or -1, 2, 2 or -2. So, 1 is an option, but the answer is 1 regardless of q being 2 or -2.But wait, let me think again. Maybe I made a mistake in the calculation.Alternatively, perhaps there's another way to compute T_7.Another approach: In a geometric sequence, the product of terms equidistant from the ends is constant.So, for example, a_1 * a_7 = a_2 * a_6 = a_3 * a_5 = a_4^2Given that, let's compute a_4.We have a_2 = 1/4, a_6 = 4.Since it's a geometric sequence, a_6 = a_2 * q^4So, 4 = (1/4) * q^4 => q^4 = 16 => q = 2 or q = -2, as before.Then, a_4 = a_2 * q^2 = (1/4) * (2)^2 = (1/4)*4 = 1Similarly, if q = -2, a_4 = (1/4)*(-2)^2 = (1/4)*4 = 1So, a_4 = 1 regardless of q.Then, the product of the first seven terms can be grouped as:(a_1 * a_7) * (a_2 * a_6) * (a_3 * a_5) * a_4Each pair like a_1 * a_7 is equal to a_4^2, which is 1^2 = 1So, we have three pairs each multiplying to 1, and then a_4 itself is 1.So, T_7 = 1 * 1 * 1 * 1 = 1Therefore, regardless of whether q is 2 or -2, T_7 is 1.So, the answer should be A: 1.Wait, but in the initial problem, the options are A:1, B:1 or -1, C:2, D:2 or -2.So, 1 is an option, so the answer is A.But wait, in my first calculation, I thought maybe it could be -1, but no, both cases gave 1.Wait, let me think again. Maybe I missed something.If q is negative, does that affect the product?Wait, in the product T_7, we have seven terms. If q is negative, then the terms alternate in sign.But since we're multiplying seven terms, which is odd, the product could be negative if an odd number of negative terms are present.But in our case, let's see.If q = 2, all terms are positive because a_1 is positive (1/8), and q is positive, so all terms are positive. So, T_7 is positive 1.If q = -2, then a_1 is negative (-1/8), and q is negative. So, the terms alternate in sign.Let's list the terms:a_1 = -1/8a_2 = (-1/8)*(-2) = 1/4a_3 = (1/4)*(-2) = -1/2a_4 = (-1/2)*(-2) = 1a_5 = 1*(-2) = -2a_6 = (-2)*(-2) = 4a_7 = 4*(-2) = -8So, the terms are: -1/8, 1/4, -1/2, 1, -2, 4, -8Now, let's compute the product:(-1/8) * (1/4) * (-1/2) * 1 * (-2) * 4 * (-8)Let's compute step by step:First, multiply (-1/8) * (1/4) = (-1/32)Then, (-1/32) * (-1/2) = (1/64)Then, (1/64) * 1 = 1/64Then, 1/64 * (-2) = (-2)/64 = (-1)/32Then, (-1)/32 * 4 = (-4)/32 = (-1)/8Then, (-1)/8 * (-8) = 1So, indeed, the product is 1.So, even though some terms are negative, the product ends up being positive 1.Therefore, regardless of q being 2 or -2, T_7 is 1.So, the answer is A:1.Wait, but the initial assistant thought process concluded T_7 = 1, but the user's initial problem had options including 1 or -1. Maybe the assistant thought it could be -1, but in reality, it's always 1.So, the correct answer is A:1.

💡Okay, so I have this problem here: the line y = -2x + m passes through points A(-2, a) and B(2, b). I need to figure out the relationship between a and b, and choose the correct option from A to D.First, let me recall what the equation y = -2x + m represents. It's a straight line in the slope-intercept form, where m is the y-intercept, and -2 is the slope. The slope tells me how steep the line is and whether it's increasing or decreasing. Since the slope is -2, which is negative, the line is decreasing. That means as x increases, y decreases, and vice versa.Now, the line passes through points A and B. Point A has coordinates (-2, a) and point B has coordinates (2, b). So, I can plug these points into the equation to find expressions for a and b.Starting with point A(-2, a):y = -2x + ma = -2*(-2) + ma = 4 + mNow, for point B(2, b):y = -2x + mb = -2*2 + mb = -4 + mSo, now I have:a = 4 + mb = -4 + mI need to find the relationship between a and b. Let's subtract the equation for b from the equation for a:a - b = (4 + m) - (-4 + m)a - b = 4 + m + 4 - ma - b = 8So, a - b = 8, which means a = b + 8. Therefore, a is greater than b by 8 units.Wait, but let me double-check. If the slope is negative, as x increases, y decreases. So, from point A(-2, a) to point B(2, b), x increases from -2 to 2, which is an increase of 4 units. Since the slope is -2, the change in y should be -2 times the change in x, so Δy = -2*4 = -8. That means y decreases by 8 units as x increases by 4 units. So, starting from a, y decreases by 8 to reach b. Therefore, a = b + 8, which confirms that a is greater than b.So, the relationship is a > b, which corresponds to option C.

💡Okay, so I have this problem where I'm given a sequence {a_n} defined by a_n = 3^n - 1 for all positive integers n. The question is asking whether there exists a geometric sequence {b_n} such that for each n, a_n can be expressed as the sum b_1c_n^1 + b_2c_n^2 + ... + b_nc_n^n. Hmm, that seems a bit abstract, but let me try to unpack it step by step.First, let's make sure I understand the notation and what's being asked. A geometric sequence {b_n} means that each term is a constant multiple of the previous one. So, if I denote the common ratio as r, then b_n = b_1 * r^{n-1} for all n. That's the definition of a geometric sequence.Now, the expression given is a_n = b_1c_n^1 + b_2c_n^2 + ... + b_nc_n^n. Wait, I think there might be a typo here because the original problem mentions c_n^k terms, but since it's a geometric sequence, maybe it's supposed to be c^k? Or perhaps c_n is a typo and it's supposed to be something else? Alternatively, maybe c_n is a constant for each n? Hmm, the problem isn't very clear on that.Wait, looking back at the problem statement, it says "a geometric sequence {b_n} such that a_n = b_1c_n^1 + b_2c_n^2 + b_3c_n^3 + ... + b_nc_n^n". So, each term in the sum is b_k multiplied by c_n^k, where k goes from 1 to n. So, for each n, we have a different c_n? Or is c_n a constant? Hmm, that's a bit confusing.Wait, maybe c_n is a typo and it's supposed to be c^k, meaning a constant c raised to the power k? That would make more sense because otherwise, if c_n is different for each n, then it's not clear how the geometric sequence {b_n} would interact with it.Alternatively, perhaps c_n is a constant for each n, but that seems less likely. Let me think. If c_n is a constant, say c, then the expression becomes a_n = b_1c^1 + b_2c^2 + ... + b_nc^n. That would make sense because then we're expressing a_n as a sum of terms from the geometric sequence {b_n} multiplied by powers of c.But in the original problem, it's written as c_n^k, which suggests that c might be a function of n. Hmm, that complicates things. Maybe it's a typo, and it's supposed to be c^k? Or perhaps c is a constant, and the subscript n is a mistake. Since the problem mentions a geometric sequence {b_n}, it's more likely that c is a constant, and the subscript n is not necessary. So, perhaps it's supposed to be c^k.Alternatively, maybe c is a constant, and c_n is a typo for c. Let me assume that for now because otherwise, the problem becomes too vague. So, let's proceed under the assumption that the expression is a_n = b_1c^1 + b_2c^2 + ... + b_nc^n, where {b_n} is a geometric sequence.Given that, let's restate the problem: Given a_n = 3^n - 1, does there exist a geometric sequence {b_n} and a constant c such that for each n, a_n is equal to the sum from k=1 to n of b_k c^k.So, in mathematical terms, we have:a_n = 3^n - 1 = sum_{k=1}^n b_k c^k, where {b_n} is a geometric sequence.Since {b_n} is a geometric sequence, we can write b_k = b_1 * r^{k-1}, where r is the common ratio.So, substituting that into the sum, we get:sum_{k=1}^n b_1 r^{k-1} c^k = b_1 c sum_{k=1}^n (r c)^{k-1}.Wait, let's see:sum_{k=1}^n b_k c^k = sum_{k=1}^n [b_1 r^{k-1}] c^k = b_1 c sum_{k=1}^n (r c)^{k-1}.That simplifies to b_1 c * sum_{k=0}^{n-1} (r c)^k.Which is a finite geometric series. The sum of a geometric series from k=0 to m is (1 - (r c)^{m+1}) / (1 - r c), provided that r c ≠ 1.So, in our case, m = n - 1, so the sum becomes (1 - (r c)^n) / (1 - r c).Therefore, the entire expression becomes:b_1 c * (1 - (r c)^n) / (1 - r c).So, we have:3^n - 1 = b_1 c * (1 - (r c)^n) / (1 - r c).Hmm, interesting. So, we have an equation that must hold for all n. Let's write that down:3^n - 1 = [b_1 c / (1 - r c)] * (1 - (r c)^n).Let me denote [b_1 c / (1 - r c)] as a constant, say K. So, we have:3^n - 1 = K (1 - (r c)^n).Our goal is to find constants K, r, and c such that this equation holds for all n.So, let's write:3^n - 1 = K - K (r c)^n.Let me rearrange this:3^n - 1 = K - K (r c)^n.Bring all terms to one side:3^n - 1 - K + K (r c)^n = 0.Hmm, so for this to hold for all n, the coefficients of like terms must be equal. Let's think about the terms involving n.We have 3^n and K (r c)^n. These are both exponential terms. For the equation to hold for all n, the bases of these exponentials must be equal, otherwise, for large n, one term would dominate and the equality would not hold.Therefore, we must have r c = 3. Because 3^n and K (r c)^n must be the same exponential term.So, set r c = 3.Then, our equation becomes:3^n - 1 = K (1 - 3^n).Wait, let's substitute r c = 3 into the equation:3^n - 1 = K (1 - 3^n).So, 3^n - 1 = K - K 3^n.Bring all terms to one side:3^n - 1 - K + K 3^n = 0.Factor out 3^n:(1 + K) 3^n - (1 + K) = 0.Wait, that's interesting. So, (1 + K)(3^n - 1) = 0.But this must hold for all n. The only way this can be true is if 1 + K = 0, because 3^n - 1 is not zero for all n (it's zero only when n=0, which is not in our consideration since n is a positive integer).So, 1 + K = 0 => K = -1.But earlier, we had K = b_1 c / (1 - r c). Since r c = 3, we have:K = b_1 c / (1 - 3).So, K = b_1 c / (-2).But we found K = -1, so:-1 = b_1 c / (-2) => Multiply both sides by (-2):2 = b_1 c.So, b_1 c = 2.But we also have r c = 3.So, we have two equations:1. r c = 32. b_1 c = 2We need to find b_1, r, and c.But we have two equations and three unknowns. So, we need another condition.Wait, but we also have that {b_n} is a geometric sequence, so b_n = b_1 r^{n-1}.But in our earlier substitution, we used that to express the sum in terms of r and c.Wait, perhaps we can find another relation from the equation when n=1.Let's test n=1.When n=1, a_1 = 3^1 - 1 = 2.On the other hand, the sum is just b_1 c^1 = b_1 c.So, 2 = b_1 c.Which is consistent with our earlier equation: b_1 c = 2.So, that doesn't give us new information.Similarly, let's test n=2.a_2 = 3^2 - 1 = 9 - 1 = 8.The sum is b_1 c^1 + b_2 c^2.But since {b_n} is geometric, b_2 = b_1 r.So, the sum is b_1 c + b_1 r c^2.We know that b_1 c = 2, so let's substitute:Sum = 2 + 2 r c.But we also have r c = 3, so:Sum = 2 + 2 * 3 = 2 + 6 = 8.Which matches a_2 = 8.Similarly, let's test n=3.a_3 = 3^3 - 1 = 27 - 1 = 26.The sum is b_1 c + b_2 c^2 + b_3 c^3.Since {b_n} is geometric, b_2 = b_1 r, b_3 = b_1 r^2.So, the sum is b_1 c + b_1 r c^2 + b_1 r^2 c^3.Factor out b_1 c:b_1 c [1 + r c + r^2 c^2].We know that b_1 c = 2, and r c = 3, so:Sum = 2 [1 + 3 + 9] = 2 * 13 = 26.Which matches a_3 = 26.Similarly, for n=4:a_4 = 3^4 - 1 = 81 - 1 = 80.Sum = b_1 c + b_2 c^2 + b_3 c^3 + b_4 c^4.Which is 2 + 6 + 18 + 54 = 80.Wait, let's compute it properly:b_1 c = 2b_2 c^2 = b_1 r c^2 = 2 * r c = 2 * 3 = 6b_3 c^3 = b_1 r^2 c^3 = 2 * (r c)^2 = 2 * 9 = 18b_4 c^4 = b_1 r^3 c^4 = 2 * (r c)^3 = 2 * 27 = 54So, sum = 2 + 6 + 18 + 54 = 80, which matches a_4.So, it seems that with b_1 c = 2 and r c = 3, the equality holds for n=1,2,3,4.Therefore, it's likely that this holds for all n.But let's try to generalize it.We have:3^n - 1 = K (1 - (r c)^n) with K = -1 and r c = 3.So, substituting K = -1 and r c = 3, we get:3^n - 1 = -1 (1 - 3^n) = -1 + 3^n.Which simplifies to 3^n - 1 = 3^n - 1, which is an identity.Therefore, the equality holds for all n.Therefore, such a geometric sequence {b_n} exists.But wait, we need to find the explicit form of {b_n}.We have:b_1 c = 2r c = 3We can choose c as a parameter, but since {b_n} is a geometric sequence, we can express it in terms of b_1 and r.But we have two equations:1. b_1 c = 22. r c = 3We can solve for b_1 and r in terms of c.From equation 1: b_1 = 2 / cFrom equation 2: r = 3 / cTherefore, the geometric sequence {b_n} is given by:b_n = b_1 r^{n-1} = (2 / c) * (3 / c)^{n-1} = 2 * 3^{n-1} / c^n.Hmm, but this seems a bit messy. Maybe we can choose c such that the expression simplifies.Wait, let's think differently. Since r c = 3, we can set c = 1, then r = 3. But then b_1 c = 2 => b_1 = 2.So, b_n = 2 * 3^{n-1}.Let's test this.For n=1: b_1 = 2 * 3^{0} = 2, which is correct.For n=2: b_2 = 2 * 3^{1} = 6For n=3: b_3 = 2 * 3^{2} = 18For n=4: b_4 = 2 * 3^{3} = 54Which matches our earlier calculations.So, with c=1, r=3, and b_1=2, the geometric sequence {b_n} is 2, 6, 18, 54, ..., which is indeed a geometric sequence with common ratio 3.Therefore, the answer is yes, such a geometric sequence exists, and it's given by b_n = 2 * 3^{n-1}.But wait, in the original problem, the sum is written as b_1 c_n^1 + b_2 c_n^2 + ... + b_n c_n^n. If c_n is a constant c for all n, then setting c=1 works, as we've done.But if c_n is different for each n, then the problem becomes more complicated. However, since in our analysis, we found that setting c=1 works, and the problem didn't specify that c_n varies with n, it's reasonable to assume that c is a constant.Therefore, the conclusion is that such a geometric sequence {b_n} exists, and it's given by b_n = 2 * 3^{n-1}.So, to summarize:Given a_n = 3^n - 1, we want to express it as a sum of terms from a geometric sequence {b_n} multiplied by powers of a constant c. By setting c=1, we find that the geometric sequence {b_n} with b_n = 2 * 3^{n-1} satisfies the condition for all n. Therefore, such a geometric sequence exists.

💡Alright, so I have this problem here. It says that the point (8, 5) is on the graph of y = g(x). I need to figure out what point must be on the graph of 3y = (g(3x))/3 + 3, and then find the sum of the coordinates of that point. Hmm, okay, let me break this down step by step.First, I know that if (8, 5) is on the graph of y = g(x), that means when x is 8, y is 5. So, in other words, g(8) = 5. Got that down. So, g(8) equals 5. That seems straightforward.Now, I need to work with the equation 3y = (g(3x))/3 + 3. I think this is a transformation of the original function g(x). Let me try to understand what transformations are happening here.Looking at the equation, on the left side, it's 3y instead of just y. That probably means there's a vertical scaling involved. On the right side, inside the function g, there's a 3x instead of just x. That suggests a horizontal scaling. Also, the entire g(3x) is divided by 3, which might be another vertical scaling, and then there's a +3 at the end, which is a vertical shift.Let me write down the equation again for clarity:3y = (g(3x))/3 + 3.I think it would help if I solve for y to make it look more like the standard function form. So, let's do that. If I divide both sides by 3, I get:y = (g(3x))/9 + 1.Wait, is that right? Let me check. If I have 3y = (g(3x))/3 + 3, then dividing both sides by 3 gives y = (g(3x))/9 + 1. Yeah, that seems correct.So, the transformed function is y = (g(3x))/9 + 1. Now, let's think about how this transformation affects the original function g(x). The 3x inside the function argument suggests a horizontal compression by a factor of 1/3. That means that the graph is squeezed horizontally, making it three times as steep. So, points on the original graph will have their x-coordinates scaled down by a factor of 3.On the other hand, the (g(3x))/9 part suggests a vertical scaling. Since it's divided by 9, that's a vertical compression by a factor of 1/9. So, the y-values of the original function are being scaled down by 1/9. Then, there's a vertical shift upwards by 1 unit because of the +1 at the end.So, putting it all together, the transformed function is a combination of horizontal compression, vertical compression, and a vertical shift. Now, I need to find a point on this transformed graph that corresponds to the point (8, 5) on the original graph.Let me denote the transformed function as h(x) = (g(3x))/9 + 1. So, h(x) = (g(3x))/9 + 1. I need to find an x such that h(x) gives me a y-value, and then I can find the corresponding point.But wait, I know that g(8) = 5. So, if I can find an x such that 3x = 8, then g(3x) would be g(8) = 5. That seems like a good approach. Let's solve for x in the equation 3x = 8.So, 3x = 8 implies that x = 8/3. Okay, so when x is 8/3, then 3x is 8, and thus g(3x) is g(8) which is 5. Perfect.Now, let's plug x = 8/3 into the transformed function h(x):h(8/3) = (g(3*(8/3)))/9 + 1 = (g(8))/9 + 1 = 5/9 + 1.Wait, 5/9 + 1 is equal to 5/9 + 9/9 = 14/9. So, h(8/3) = 14/9. That means the point (8/3, 14/9) is on the transformed graph.Now, the question asks for the sum of the coordinates of this point. So, I need to add 8/3 and 14/9. To add these, they need a common denominator. The denominators are 3 and 9, so the common denominator is 9.Converting 8/3 to ninths: 8/3 = 24/9. So, 24/9 + 14/9 = 38/9.Therefore, the sum of the coordinates is 38/9.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning, (8,5) is on y = g(x), so g(8) = 5. Then, the transformed equation is 3y = (g(3x))/3 + 3. Solving for y gives y = (g(3x))/9 + 1. To find a corresponding point, set 3x = 8, so x = 8/3. Plugging back in, y = (g(8))/9 + 1 = 5/9 + 1 = 14/9. So, the point is (8/3, 14/9). Adding 8/3 and 14/9: 8/3 is 24/9, so 24/9 + 14/9 = 38/9. Yep, that seems correct.I think I've got it. The sum of the coordinates is 38/9.**Final Answer**The sum of the coordinates of the point is boxed{dfrac{38}{9}}.