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∑n=1∞n2+15n\sum_{n=1}^{\infty}\frac{n^{2}+1}{5^{n}}

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Q. ∑n=1∞n2+15n\sum_{n=1}^{\infty}\frac{n^{2}+1}{5^{n}}
  1. Recognize Series Components: We are given the series ∑n=1∞n2+15n\sum_{n=1}^{\infty}\frac{n^{2}+1}{5^{n}}. To find the sum of this series, we need to recognize that it is not a geometric series, but it can be broken down into two separate series: one that is geometric and one that can be solved using power series techniques.
  2. Calculate Geometric Series Sum: First, let's separate the series into two parts: ∑n=1∞n25n\sum_{n=1}^{\infty}\frac{n^{2}}{5^{n}} and ∑n=1∞15n\sum_{n=1}^{\infty}\frac{1}{5^{n}}. The second series is a geometric series with a common ratio of 15\frac{1}{5} and the first term a1=15a_1 = \frac{1}{5}.
  3. Use Power Series Techniques: The sum of the geometric series ∑n=1∞15n\sum_{n=1}^{\infty}\frac{1}{5^{n}} is given by a11−r\frac{a_1}{1 - r}, where a1a_1 is the first term and rr is the common ratio. Plugging in the values, we get 1/51−1/5=1/54/5=14\frac{1/5}{1 - 1/5} = \frac{1/5}{4/5} = \frac{1}{4}.
  4. Differentiate Known Geometric Series: Now, we need to find the sum of the series ∑n=1∞n25n\sum_{n=1}^{\infty}\frac{n^{2}}{5^{n}}. This is not a geometric series, but we can use the fact that the sum of the series ∑n=1∞xnns\sum_{n=1}^{\infty}\frac{x^n}{n^s} is related to the polylogarithm function, which is beyond the scope of elementary calculus. However, we can differentiate a known geometric series to find a related series.
  5. Substitute xx into Derived Series: Consider the geometric series ∑n=0∞xn=11−x\sum_{n=0}^{\infty}x^n = \frac{1}{1-x} for |x| < 1. Differentiating both sides with respect to xx gives ∑n=1∞nxn−1=1(1−x)2\sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}. We can then multiply both sides by xx to get ∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty}nx^n = \frac{x}{(1-x)^2}.
  6. Simplify Expression: Differentiating both sides of ∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty}nx^n = \frac{x}{(1-x)^2} with respect to xx again gives ∑n=1∞n2xn−1=1+x(1−x)3\sum_{n=1}^{\infty}n^2x^{n-1} = \frac{1+x}{(1-x)^3}. Multiplying both sides by xx, we get ∑n=1∞n2xn=x(1+x)(1−x)3\sum_{n=1}^{\infty}n^2x^n = \frac{x(1+x)}{(1-x)^3}.
  7. Combine Geometric and Derived Series: Now, we substitute x=15x = \frac{1}{5} into the derived series ∑n=1∞n2xn=x(1+x)(1−x)3\sum_{n=1}^{\infty}n^2x^n = x\frac{(1+x)}{(1-x)^3} to find the sum of ∑n=1∞n25n\sum_{n=1}^{\infty}\frac{n^{2}}{5^{n}}. We get ∑n=1∞n25n=15(1+15)(1−15)3=1565(45)3\sum_{n=1}^{\infty}\frac{n^{2}}{5^{n}} = \frac{1}{5}\frac{(1+\frac{1}{5})}{(1-\frac{1}{5})^3} = \frac{1}{5}\frac{\frac{6}{5}}{\left(\frac{4}{5}\right)^3}.
  8. Find Total Sum: Simplify the expression (15)(65)/(45)3(\frac{1}{5})(\frac{6}{5})/(\frac{4}{5})^3. This simplifies to (625)/(64125)=(625)⋅(12564)=(6⋅12525⋅64)=7501600=1532(\frac{6}{25})/(\frac{64}{125}) = (\frac{6}{25}) \cdot (\frac{125}{64}) = (\frac{6 \cdot 125}{25 \cdot 64}) = \frac{750}{1600} = \frac{15}{32}.
  9. Final Answer: Now, we add the sum of the geometric series to the sum of the series involving n2n^2. The total sum is 14+1532\frac{1}{4} + \frac{15}{32}. To combine these, we need a common denominator, which is 3232. So we get (832)+(1532)=2332(\frac{8}{32}) + (\frac{15}{32}) = \frac{23}{32}.
  10. Final Answer: Now, we add the sum of the geometric series to the sum of the series involving n2n^2. The total sum is 14+1532\frac{1}{4} + \frac{15}{32}. To combine these, we need a common denominator, which is 3232. So we get (832)+(1532)=2332(\frac{8}{32}) + (\frac{15}{32}) = \frac{23}{32}.The final answer is the sum of the series ∑n=1∞(n2+1)/(5n)=2332\sum_{n=1}^{\infty}(n^{2}+1)/(5^{n}) = \frac{23}{32}.

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