💡Alright, so I have this problem about a circle and a point outside of it. Let me try to figure it out step by step. First, the problem says that point P(0, 4) is outside the circle C defined by the equation (x^2 + y^2 + 6x - 8y + m = 0). There are three parts to this problem, and I need to solve each one.Starting with part (1): Find the range of the real number (m).Hmm, okay. I remember that for a point to be outside a circle, the distance from the point to the center of the circle must be greater than the radius of the circle. So, I need to find the center and the radius of the given circle first.The general equation of a circle is (x^2 + y^2 + Dx + Ey + F = 0), and its center is at ((-D/2, -E/2)) with radius (sqrt{(D/2)^2 + (E/2)^2 - F}). Let me apply this to the given equation.Here, (D = 6), (E = -8), and (F = m). So, the center of the circle should be at ((-6/2, -(-8)/2)), which simplifies to ((-3, 4)). Now, the radius (r) is calculated as (sqrt{(6/2)^2 + (-8/2)^2 - m}). Calculating that, it's (sqrt{3^2 + (-4)^2 - m}), which is (sqrt{9 + 16 - m}) or (sqrt{25 - m}).Wait, hold on. The radius must be a real number, so the expression inside the square root must be positive. That means (25 - m > 0), so (m < 25). That's one condition.Next, since point P(0, 4) is outside the circle, the distance from P to the center must be greater than the radius. Let's compute the distance between P(0, 4) and the center (-3, 4). The distance formula is (sqrt{(x2 - x1)^2 + (y2 - y1)^2}). Plugging in the values, it's (sqrt{(-3 - 0)^2 + (4 - 4)^2}) which simplifies to (sqrt{9 + 0}) or 3. So, the distance from P to the center is 3. This must be greater than the radius (r = sqrt{25 - m}). Therefore, we have (3 > sqrt{25 - m}). To solve this inequality, I'll square both sides: (9 > 25 - m). Then, rearranging, (m > 25 - 9), which is (m > 16).So, combining both conditions, (m) must be greater than 16 and less than 25. Therefore, the range of (m) is (16 < m < 25). Wait, let me double-check that. If (m = 16), the radius would be (sqrt{25 - 16} = 3), which is equal to the distance from P to the center. That would mean P is on the circle, but the problem states it's outside. So, (m) must be strictly greater than 16. Similarly, if (m = 25), the radius would be zero, which isn't a circle anymore. So, (m) must be less than 25. Yep, that seems right.Moving on to part (2): If (m = 24), find the minimum value of (x^2 + y^2).Alright, so (m = 24). Let me write the equation of the circle again with this value. The equation becomes (x^2 + y^2 + 6x - 8y + 24 = 0). I can rewrite this in standard form by completing the squares. Let's group the x terms and y terms:(x^2 + 6x + y^2 - 8y = -24).Completing the square for x: Take the coefficient of x, which is 6, divide by 2 to get 3, square it to get 9. Similarly, for y: Coefficient is -8, divide by 2 to get -4, square it to get 16.So, adding these to both sides:(x^2 + 6x + 9 + y^2 - 8y + 16 = -24 + 9 + 16).Simplifying, that's ((x + 3)^2 + (y - 4)^2 = 1).So, the circle has center at (-3, 4) and radius 1.Now, the problem asks for the minimum value of (x^2 + y^2). Hmm, (x^2 + y^2) is the square of the distance from any point (x, y) to the origin (0, 0). So, we're looking for the minimum distance squared from the origin to a point on the circle.To find this, I can visualize the origin and the circle. The minimum distance from the origin to the circle will be the distance from the origin to the center of the circle minus the radius.First, let's compute the distance from the origin (0, 0) to the center (-3, 4). Using the distance formula: (sqrt{(-3 - 0)^2 + (4 - 0)^2}) which is (sqrt{9 + 16}) or (sqrt{25}), which is 5.So, the distance from the origin to the center is 5. The radius of the circle is 1. Therefore, the minimum distance from the origin to the circle is (5 - 1 = 4). Hence, the minimum value of (x^2 + y^2) is (4^2 = 16).Wait, let me make sure. Is it possible that the origin is inside the circle? If the distance from the origin to the center is 5, and the radius is 1, then the origin is outside the circle because 5 > 1. So, the minimum distance is indeed 5 - 1 = 4, and the minimum value of (x^2 + y^2) is 16. That seems correct.Now, part (3): Under the condition of question (2), find the range of (frac{y - 4}{x}).So, with (m = 24), the circle is ((x + 3)^2 + (y - 4)^2 = 1). We need to find the range of (frac{y - 4}{x}).Let me denote (k = frac{y - 4}{x}). Then, this can be rewritten as (y = kx + 4). So, we have the line (y = kx + 4) intersecting the circle. For this line to intersect the circle, the distance from the center of the circle to the line must be less than or equal to the radius.The center of the circle is (-3, 4). Let's compute the distance from (-3, 4) to the line (y = kx + 4).The formula for the distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is (frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}).First, let's write the line (y = kx + 4) in standard form: (kx - y + 4 = 0). So, A = k, B = -1, C = 4.Plugging into the distance formula:Distance = (frac{|k*(-3) + (-1)*4 + 4|}{sqrt{k^2 + (-1)^2}})Simplify numerator:(|-3k - 4 + 4| = |-3k| = |3k|)Denominator:(sqrt{k^2 + 1})So, the distance is (frac{3|k|}{sqrt{k^2 + 1}}).Since the line must intersect the circle, this distance must be less than or equal to the radius, which is 1.So, (frac{3|k|}{sqrt{k^2 + 1}} leq 1)Let me solve this inequality:Multiply both sides by (sqrt{k^2 + 1}):(3|k| leq sqrt{k^2 + 1})Square both sides to eliminate the square root:(9k^2 leq k^2 + 1)Subtract (k^2) from both sides:(8k^2 leq 1)Divide both sides by 8:(k^2 leq frac{1}{8})Take square roots:(|k| leq frac{1}{2sqrt{2}} = frac{sqrt{2}}{4})So, (k) is between (-frac{sqrt{2}}{4}) and (frac{sqrt{2}}{4}).Therefore, the range of (frac{y - 4}{x}) is (left[ -frac{sqrt{2}}{4}, frac{sqrt{2}}{4} right]).Let me verify this. If I consider the line (y = kx + 4) tangent to the circle, then the maximum and minimum values of (k) would occur when the line is tangent, which is exactly when the distance equals the radius. So, solving for equality gives the boundary values, which are (pm frac{sqrt{2}}{4}). That makes sense.So, summarizing:1. The range of (m) is (16, 25).2. The minimum value of (x^2 + y^2) is 16.3. The range of (frac{y - 4}{x}) is (left[ -frac{sqrt{2}}{4}, frac{sqrt{2}}{4} right]).I think that covers all parts of the problem.**Final Answer**(1) The range of (m) is (boxed{(16, 25)}).(2) The minimum value of (x^2 + y^2) is (boxed{16}).(3) The range of (frac{y - 4}{x}) is (boxed{left[ -frac{sqrt{2}}{4}, frac{sqrt{2}}{4} right]}).

💡Okay, so I have this problem where there are four points A, B, C, D on the surface of a sphere. The segments AB, AC, and AD are mutually perpendicular, and their sum is 12. I need to find the minimum surface area of the sphere. Hmm, let me try to visualize this.First, since AB, AC, and AD are mutually perpendicular, it reminds me of the edges of a rectangular box meeting at a corner. So, if I imagine point A as one corner of a box, then B, C, and D would be the adjacent corners connected by edges AB, AC, and AD respectively. That makes sense because those edges are perpendicular to each other.Now, since all these points are on the surface of a sphere, the sphere must contain this entire box. The sphere's diameter would then be the space diagonal of the box. The space diagonal of a rectangular box with sides a, b, c is given by the formula sqrt(a² + b² + c²). So, the radius of the sphere would be half of that, which is (1/2) * sqrt(a² + b² + c²).The surface area of a sphere is 4πr². So, if I can find the radius, I can compute the surface area. Let me write that down:Surface Area = 4π * [(1/2) * sqrt(a² + b² + c²)]²= 4π * (1/4) * (a² + b² + c²)= π * (a² + b² + c²)So, the surface area is π times the sum of the squares of the edges AB, AC, and AD.Now, the problem states that AB + AC + AD = 12. Let me denote AB = a, AC = b, and AD = c. So, a + b + c = 12.I need to minimize the surface area, which is π(a² + b² + c²). To minimize this, I should minimize the sum of the squares given that the sum of a, b, c is fixed at 12.I remember from algebra that for a fixed sum, the sum of squares is minimized when all the variables are equal. This is related to the Cauchy-Schwarz inequality or the concept of minimizing variance. So, if a = b = c, then the sum of squares will be minimized.Let me test this. If a = b = c, then each would be 12/3 = 4. So, a = b = c = 4.Then, the sum of squares would be 4² + 4² + 4² = 16 + 16 + 16 = 48.Therefore, the surface area would be π * 48 = 48π.Is this the minimum? Let me think. Suppose I take different values for a, b, c that still add up to 12. For example, let me take a = 5, b = 5, c = 2. Then, the sum of squares is 25 + 25 + 4 = 54, which is larger than 48. Another example: a = 6, b = 3, c = 3. Sum of squares is 36 + 9 + 9 = 54 again. So, it seems that when the variables are unequal, the sum of squares increases.Therefore, the minimal sum of squares occurs when a = b = c = 4, leading to a surface area of 48π.Wait, but let me double-check if there's a possibility of a smaller sum of squares. Suppose I take a = 0, but then the other two would have to add up to 12. But if a = 0, then b + c = 12, and the sum of squares would be 0 + b² + c². The minimal sum of squares for two variables with a fixed sum occurs when they are equal, so b = c = 6. Then, sum of squares is 0 + 36 + 36 = 72, which is way larger than 48. So, definitely, setting a = 0 isn't helpful.Alternatively, what if I take a = 3, b = 4, c = 5? Then, sum of squares is 9 + 16 + 25 = 50, which is still larger than 48. So, 48 seems to be the minimal.Therefore, the minimal surface area is 48π, which corresponds to option C.**Final Answer**boxed{C}

💡Okay, so I have this problem here: In the expansion of the binomial ((ax + frac{1}{x})^6), if the constant term is (-160), then what is the value of (a)? Hmm, let me try to figure this out step by step.First, I remember that when you expand a binomial like ((A + B)^n), each term in the expansion is given by the binomial theorem. The general term is (C(n, k) cdot A^{n-k} cdot B^k), where (C(n, k)) is the combination of (n) things taken (k) at a time. So in this case, (A) is (ax) and (B) is (frac{1}{x}), and (n) is 6.So, the general term in the expansion of ((ax + frac{1}{x})^6) would be (C(6, k) cdot (ax)^{6 - k} cdot left(frac{1}{x}right)^k). Let me write that out:(T_{k+1} = C(6, k) cdot (ax)^{6 - k} cdot left(frac{1}{x}right)^k).Now, I need to find the constant term. A constant term is the term where the exponent of (x) is zero because (x^0 = 1), which is a constant. So, I need to find the value of (k) such that the exponent of (x) in the term (T_{k+1}) is zero.Let me simplify the term (T_{k+1}):First, ((ax)^{6 - k}) is (a^{6 - k} cdot x^{6 - k}).Then, (left(frac{1}{x}right)^k) is (x^{-k}).Multiplying these together, the term becomes:(C(6, k) cdot a^{6 - k} cdot x^{6 - k} cdot x^{-k}).Combine the exponents of (x):(x^{6 - k - k} = x^{6 - 2k}).So, the exponent of (x) in the term is (6 - 2k). We want this exponent to be zero for the term to be constant. Therefore:(6 - 2k = 0).Solving for (k):(6 = 2k)(k = 3).Okay, so the constant term is the term when (k = 3). Let me write that term:(T_{4} = C(6, 3) cdot a^{6 - 3} cdot x^{6 - 2 cdot 3}).Simplify:(C(6, 3)) is the combination of 6 things taken 3 at a time. I remember that (C(6, 3) = 20). Let me verify that:(C(6, 3) = frac{6!}{3! cdot (6 - 3)!} = frac{720}{6 cdot 6} = frac{720}{36} = 20). Yeah, that's correct.So, (T_{4} = 20 cdot a^{3} cdot x^{0}), which simplifies to (20a^3).But the problem says that the constant term is (-160). So:(20a^3 = -160).Now, I need to solve for (a). Let's divide both sides by 20:(a^3 = frac{-160}{20})(a^3 = -8).To find (a), I take the cube root of both sides:(a = sqrt[3]{-8}).I know that the cube root of -8 is -2 because ((-2)^3 = -8).So, (a = -2).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded the binomial using the general term: correct.2. Calculated the exponent of (x) correctly: (6 - 2k).3. Set the exponent equal to zero and solved for (k): got (k = 3), correct.4. Calculated the combination (C(6, 3)): 20, correct.5. Plugged into the term: (20a^3), correct.6. Set equal to -160: (20a^3 = -160), correct.7. Solved for (a^3 = -8), then (a = -2): correct.Everything seems to check out. So, I think the value of (a) is indeed -2.**Final Answer**The value of (a) is boxed{-2}.

💡Okay, so I have this problem where there's a sequence defined by a recurrence relation. The sequence starts with a₀ = 3 and a₁ = 2, and then each term after that is the sum of the two previous terms. So, it's similar to the Fibonacci sequence but starting with different initial values.The problem is asking me to find the sum from n = 0 to 8 of aₙ divided by (aₙ₊₁ times aₙ₊₂). That is, I need to compute:[ sum_{n=0}^{8} frac{a_n}{a_{n+1} a_{n+2}} ]Alright, so first, let me write down what I know. The sequence is defined by:- a₀ = 3- a₁ = 2- aₙ₊₂ = aₙ₊₁ + aₙ for all n ≥ 0So, this is a linear recurrence relation of order 2. The terms will grow similarly to the Fibonacci sequence but starting from different points.Now, the sum I need to compute involves terms of the form aₙ / (aₙ₊₁ aₙ₊₂). Hmm, that looks a bit complicated, but maybe I can find a way to simplify it.I remember that sometimes fractions like this can be expressed as telescoping series. A telescoping series is one where most terms cancel out when you add them up, leaving only the first and last terms. To see if that's possible here, I should try to express the fraction aₙ / (aₙ₊₁ aₙ₊₂) as the difference of two simpler fractions.Let me think about how to manipulate the fraction:[ frac{a_n}{a_{n+1} a_{n+2}} ]Given the recurrence relation, aₙ₊₂ = aₙ₊₁ + aₙ. So, I can substitute that into the denominator:[ frac{a_n}{a_{n+1} (a_{n+1} + a_n)} ]Hmm, that might help. Let me see if I can split this into partial fractions. Maybe something like:[ frac{a_n}{a_{n+1} (a_{n+1} + a_n)} = frac{A}{a_{n+1}} + frac{B}{a_{n+1} + a_n} ]But I'm not sure if that's the right approach. Alternatively, maybe I can express it as a difference of reciprocals. Let me try that.Since aₙ₊₂ = aₙ₊₁ + aₙ, then aₙ = aₙ₊₂ - aₙ₊₁. So, substituting that into the numerator:[ frac{a_n}{a_{n+1} a_{n+2}} = frac{a_{n+2} - a_{n+1}}{a_{n+1} a_{n+2}} ]Now, I can split this into two fractions:[ frac{a_{n+2}}{a_{n+1} a_{n+2}} - frac{a_{n+1}}{a_{n+1} a_{n+2}} ]Simplifying each term:[ frac{1}{a_{n+1}} - frac{1}{a_{n+2}} ]Oh! That's a telescoping series! So, each term in the sum can be written as the difference of two reciprocals. That means when I add them up, most terms will cancel out.So, the sum becomes:[ sum_{n=0}^{8} left( frac{1}{a_{n+1}} - frac{1}{a_{n+2}} right) ]Let me write out the terms to see the cancellation:- For n = 0: 1/a₁ - 1/a₂- For n = 1: 1/a₂ - 1/a₃- For n = 2: 1/a₃ - 1/a₄- ...- For n = 8: 1/a₉ - 1/a₁₀When I add all these up, the -1/a₂ from the first term cancels with the +1/a₂ from the second term, the -1/a₃ cancels with the +1/a₃ from the third term, and so on. This pattern continues, and all the intermediate terms cancel out. What's left is the first term of the first expression and the last term of the last expression.So, the sum simplifies to:[ frac{1}{a_1} - frac{1}{a_{10}} ]Now, I need to compute this. I know that a₁ = 2, so 1/a₁ is 1/2. But I need to find a₁₀. To do that, I can compute the terms of the sequence up to a₁₀ using the recurrence relation.Let me list out the terms step by step:- a₀ = 3- a₁ = 2- a₂ = a₁ + a₀ = 2 + 3 = 5- a₃ = a₂ + a₁ = 5 + 2 = 7- a₄ = a₃ + a₂ = 7 + 5 = 12- a₅ = a₄ + a₃ = 12 + 7 = 19- a₆ = a₅ + a₄ = 19 + 12 = 31- a₇ = a₆ + a₅ = 31 + 19 = 50- a₈ = a₇ + a₆ = 50 + 31 = 81- a₉ = a₈ + a₇ = 81 + 50 = 131- a₁₀ = a₉ + a₈ = 131 + 81 = 212So, a₁₀ is 212. Therefore, 1/a₁₀ is 1/212.Now, plugging back into the sum:[ frac{1}{a_1} - frac{1}{a_{10}} = frac{1}{2} - frac{1}{212} ]To subtract these fractions, I need a common denominator. The least common denominator of 2 and 212 is 212.Converting 1/2 to 212ths:[ frac{1}{2} = frac{106}{212} ]So, subtracting:[ frac{106}{212} - frac{1}{212} = frac{105}{212} ]Therefore, the sum is 105/212.Let me double-check my calculations to make sure I didn't make any mistakes. Starting from the sequence:- a₀ = 3- a₁ = 2- a₂ = 5- a₃ = 7- a₄ = 12- a₅ = 19- a₆ = 31- a₇ = 50- a₈ = 81- a₉ = 131- a₁₀ = 212Yes, that seems correct. Each term is the sum of the two before it. So, a₁₀ is indeed 212.Then, the sum simplifies to 1/2 - 1/212, which is 105/212. That seems right.I think I've got it. The sum is 105/212.

💡Okay, so I have this equation: ( x^2 + y^2 + 2x - 4y - 8 = 3y - 6x + 9 ). Hmm, I need to find the area of the region defined by this equation. I remember that equations involving ( x^2 ) and ( y^2 ) usually represent circles, ellipses, or hyperbolas, but since both ( x ) and ( y ) are squared and have the same sign, it's probably a circle or an ellipse. Let me see.First, I think I should simplify the equation by moving all terms to one side. So, I'll subtract ( 3y ) and add ( 6x ) and subtract 9 from both sides to get everything on the left. Let me write that out:( x^2 + y^2 + 2x - 4y - 8 - 3y + 6x - 9 = 0 )Now, let me combine like terms. For the ( x ) terms: ( 2x + 6x = 8x ). For the ( y ) terms: ( -4y - 3y = -7y ). And the constants: ( -8 - 9 = -17 ). So the equation simplifies to:( x^2 + y^2 + 8x - 7y - 17 = 0 )Alright, this looks like the general form of a circle equation, which is ( x^2 + y^2 + Dx + Ey + F = 0 ). To find the area, I need to find the radius of the circle. To do that, I should complete the square for both ( x ) and ( y ) terms.Starting with the ( x ) terms: ( x^2 + 8x ). To complete the square, I take half of the coefficient of ( x ), which is 8, so half of that is 4, and then square it, which gives 16. So, ( x^2 + 8x = (x + 4)^2 - 16 ).Now for the ( y ) terms: ( y^2 - 7y ). Similarly, take half of -7, which is -3.5 or ( -frac{7}{2} ), and square it, which gives ( frac{49}{4} ). So, ( y^2 - 7y = left(y - frac{7}{2}right)^2 - frac{49}{4} ).Now, substitute these back into the equation:( (x + 4)^2 - 16 + left(y - frac{7}{2}right)^2 - frac{49}{4} - 17 = 0 )Combine the constants: -16 - ( frac{49}{4} ) - 17. Let me convert them all to quarters to combine easily. -16 is ( -frac{64}{4} ), -17 is ( -frac{68}{4} ), so total is ( -frac{64}{4} - frac{49}{4} - frac{68}{4} = -frac{181}{4} ).So, the equation becomes:( (x + 4)^2 + left(y - frac{7}{2}right)^2 - frac{181}{4} = 0 )Move the constant term to the other side:( (x + 4)^2 + left(y - frac{7}{2}right)^2 = frac{181}{4} )So, this is the standard form of a circle with center at ( (-4, frac{7}{2}) ) and radius squared ( frac{181}{4} ). Therefore, the radius is ( sqrt{frac{181}{4}} = frac{sqrt{181}}{2} ).The area of a circle is ( pi r^2 ), so plugging in the radius squared:Area = ( pi times frac{181}{4} = frac{181pi}{4} )Wait a second, let me double-check my calculations because earlier I thought the radius squared was ( frac{153}{4} ). Did I make a mistake?Looking back, when I combined the constants after completing the square, I had:-16 (from ( x ) terms) - ( frac{49}{4} ) (from ( y ) terms) -17 (constant term). So that's -16 -17 = -33, and -33 - ( frac{49}{4} ). Converting -33 to quarters: -33 = -( frac{132}{4} ), so total is -( frac{132}{4} - frac{49}{4} = -frac{181}{4} ). So that part is correct.Therefore, the radius squared is ( frac{181}{4} ), so the area is ( frac{181pi}{4} ).But wait, in my initial thought process, I thought the radius squared was ( frac{153}{4} ). Maybe I confused it with another problem. Let me check the steps again.Original equation after moving terms: ( x^2 + y^2 + 8x - 7y - 17 = 0 ). Completing the square:( x^2 + 8x = (x + 4)^2 - 16 )( y^2 - 7y = left(y - frac{7}{2}right)^2 - frac{49}{4} )Substituting back:( (x + 4)^2 - 16 + left(y - frac{7}{2}right)^2 - frac{49}{4} - 17 = 0 )Combine constants: -16 - ( frac{49}{4} ) -17 = -16 -17 - ( frac{49}{4} ) = -33 - ( frac{49}{4} ) = -( frac{132}{4} - frac{49}{4} ) = -( frac{181}{4} )So, equation is ( (x + 4)^2 + left(y - frac{7}{2}right)^2 = frac{181}{4} ). Therefore, radius squared is ( frac{181}{4} ), so area is ( frac{181pi}{4} ).I think my initial thought of ( frac{153pi}{4} ) was incorrect. The correct area should be ( frac{181pi}{4} ).Wait, but in the initial problem, the equation was ( x^2 + y^2 + 2x - 4y - 8 = 3y - 6x + 9 ). Let me re-express that correctly.Starting over:Original equation: ( x^2 + y^2 + 2x - 4y - 8 = 3y - 6x + 9 )Bring all terms to the left:( x^2 + y^2 + 2x - 4y - 8 - 3y + 6x - 9 = 0 )Combine like terms:( x^2 + (2x + 6x) + y^2 + (-4y - 3y) + (-8 - 9) = 0 )Which is:( x^2 + 8x + y^2 - 7y - 17 = 0 )So that's correct. Then completing the square:( x^2 + 8x = (x + 4)^2 - 16 )( y^2 - 7y = left(y - frac{7}{2}right)^2 - frac{49}{4} )Substituting back:( (x + 4)^2 - 16 + left(y - frac{7}{2}right)^2 - frac{49}{4} - 17 = 0 )Combine constants:-16 - ( frac{49}{4} ) -17 = -16 -17 - ( frac{49}{4} ) = -33 - ( frac{49}{4} ) = -( frac{132}{4} - frac{49}{4} ) = -( frac{181}{4} )Thus, equation becomes:( (x + 4)^2 + left(y - frac{7}{2}right)^2 = frac{181}{4} )So radius squared is ( frac{181}{4} ), area is ( frac{181pi}{4} ).I think I made a mistake earlier when I thought it was 153. Maybe I confused it with another problem. So, the correct area is ( frac{181pi}{4} ).

💡Okay, so I have this problem where I need to compare the sum (frac{1}{2!} + frac{2}{3!} + ldots + frac{2012}{2013!}) with 1. Hmm, factorials can get pretty big, so I wonder how this sum behaves. Let me try to break it down step by step.First, let me write out the general term of the series. It seems like each term is of the form (frac{n}{(n+1)!}) where (n) starts from 1 and goes up to 2012. So, the series can be written as:[sum_{n=1}^{2012} frac{n}{(n+1)!}]I think I remember that factorials grow very rapidly, so each term in the series should get smaller as (n) increases. But I need to figure out if the entire sum is less than, equal to, or greater than 1.Maybe I can find a pattern or a way to simplify each term. Let me look at the term (frac{n}{(n+1)!}). I know that ((n+1)! = (n+1) times n!), so I can rewrite the term as:[frac{n}{(n+1)!} = frac{n}{(n+1) times n!} = frac{1}{(n+1) times (n-1)!}]Wait, that doesn't seem immediately helpful. Maybe there's another way to express (frac{n}{(n+1)!}). Let me think about telescoping series because sometimes those can collapse nicely.I recall that sometimes fractions can be expressed as differences of other fractions, which allows terms to cancel out when summed. Let me see if I can write (frac{n}{(n+1)!}) as a difference.Let me try:[frac{n}{(n+1)!} = frac{(n+1) - 1}{(n+1)!} = frac{n+1}{(n+1)!} - frac{1}{(n+1)!}]Simplifying each part:[frac{n+1}{(n+1)!} = frac{1}{n!}]and[frac{1}{(n+1)!} = frac{1}{(n+1)!}]So, putting it together:[frac{n}{(n+1)!} = frac{1}{n!} - frac{1}{(n+1)!}]Oh, that's a useful decomposition! So each term in the series can be written as the difference of two reciprocal factorials. Let me substitute this back into the original series:[sum_{n=1}^{2012} left( frac{1}{n!} - frac{1}{(n+1)!} right)]Now, when I expand this sum, it should telescope, meaning most terms will cancel out. Let me write out the first few terms to see the pattern:For (n=1):[frac{1}{1!} - frac{1}{2!}]For (n=2):[frac{1}{2!} - frac{1}{3!}]For (n=3):[frac{1}{3!} - frac{1}{4!}]...For (n=2012):[frac{1}{2012!} - frac{1}{2013!}]When I add all these up, let's see what happens:[left( frac{1}{1!} - frac{1}{2!} right) + left( frac{1}{2!} - frac{1}{3!} right) + left( frac{1}{3!} - frac{1}{4!} right) + ldots + left( frac{1}{2012!} - frac{1}{2013!} right)]Notice that each negative term cancels with the positive term in the next fraction. Specifically, (-frac{1}{2!}) cancels with (+frac{1}{2!}), (-frac{1}{3!}) cancels with (+frac{1}{3!}), and so on, all the way up to (-frac{1}{2012!}) canceling with (+frac{1}{2012!}).After all these cancellations, the only terms that remain are the very first term and the very last term:[frac{1}{1!} - frac{1}{2013!}]Simplifying this:[1 - frac{1}{2013!}]So, the entire sum simplifies to (1 - frac{1}{2013!}). Now, since (2013!) is a very large number, (frac{1}{2013!}) is a very small positive number. Therefore, (1 - frac{1}{2013!}) is just a tiny bit less than 1.Hence, the sum (frac{1}{2!} + frac{2}{3!} + ldots + frac{2012}{2013!}) is less than 1.**Final Answer**The number (frac{1}{2!} + frac{2}{3!} + ldots + frac{2012}{2013!}) is less than 1.(boxed{text{First number is smaller}})