Question

The sum of the infinite series, $1^2 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \frac{5^2}{5^4} - \frac{6^2}{5^5} + \dots$ is :

• A. $\frac{1}{2}$
• B. $\frac{25}{24}$
• C. $\frac{25}{54}$
• D. $\frac{125}{252}$

Answer 1

Answer: (C)

$\frac{25}{54}$

Answer 2

The given infinite series is $S = 1^2 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \frac{5^2}{5^4} - \frac{6^2}{5^5} + \dots$.

We can write the general term of the series as $T_n = (-1)^{n-1} \frac{n^2}{5^{n-1}}$. So, the series can be expressed in summation notation as: $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n^2}{5^{n-1}} = \sum_{n=1}^{\infty} n^2 \left(-\frac{1}{5}\right)^{n-1}$.

Let $x = -\frac{1}{5}$. Since $|x| = \left|-\frac{1}{5}\right| = \frac{1}{5} < 1$, the series converges. The series becomes $S = \sum_{n=1}^{\infty} n^2 x^{n-1}$.

We know the sum of the geometric series for $|x|<1$: $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ Differentiating both sides with respect to $x$: $\sum_{n=1}^{\infty} n x^{n-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}$ Let this sum be $S_1 = \frac{1}{(1-x)^2}$. Now, multiply both sides by $x$: $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$ Let this sum be $S_2 = \frac{x}{(1-x)^2}$. Differentiating both sides of $S_2$ with respect to $x$ again: $\sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{d}{dx}\left(\frac{x}{(1-x)^2}\right)$ Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u=x$ and $v=(1-x)^2$: $u' = \frac{d}{dx}(x) = 1$ $v' = \frac{d}{dx}((1-x)^2) = 2(1-x)(-1) = -2(1-x)$ So, $\sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{1 \cdot (1-x)^2 - x \cdot (-2(1-x))}{\left((1-x)^2\right)^2}$ $= \frac{(1-x)^2 + 2x(1-x)}{(1-x)^4}$ Factor out $(1-x)$ from the numerator: $= \frac{(1-x)( (1-x) + 2x )}{(1-x)^4}$ $= \frac{(1-x)(1+x)}{(1-x)^4}$ $= \frac{1+x}{(1-x)^3}$ Now, substitute $x = -\frac{1}{5}$ into this formula: $S = \frac{1 + \left(-\frac{1}{5}\right)}{\left(1 - \left(-\frac{1}{5}\right)\right)^3}$ $S = \frac{1 - \frac{1}{5}}{\left(1 + \frac{1}{5}\right)^3}$ $S = \frac{\frac{5-1}{5}}{\left(\frac{5+1}{5}\right)^3}$ $S = \frac{\frac{4}{5}}{\left(\frac{6}{5}\right)^3}$ $S = \frac{\frac{4}{5}}{\frac{6^3}{5^3}}$ $S = \frac{4}{5} \cdot \frac{5^3}{6^3}$ $S = \frac{4 \cdot 5^2}{6^3}$ $S = \frac{4 \cdot 25}{216}$ To simplify the fraction, divide the numerator and denominator by their greatest common divisor, which is 4: $S = \frac{100 \div 4}{216 \div 4} = \frac{25}{54}$

The final answer is $\frac{25}{54}$.

Explanation of the solution:

1. Identify the series form: The given series $1^2 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \dots$ is an infinite series of the form $\sum_{n=1}^{\infty} n^2 x^{n-1}$, where $x = -\frac{1}{5}$.
2. Recall/Derive power series sum: Start with the geometric series sum $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$.
3. Differentiate twice:
   • Differentiate once with respect to $x$: $\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$.
   • Multiply by $x$: $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$.
   • Differentiate again with respect to $x$: $\sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{d}{dx}\left(\frac{x}{(1-x)^2}\right)$.
4. Apply quotient rule: Using the quotient rule, the derivative simplifies to $\frac{(1-x)(1+x)}{(1-x)^4} = \frac{1+x}{(1-x)^3}$.
5. Substitute value of x: Substitute $x = -\frac{1}{5}$ into the derived formula: $S = \frac{1 + (-\frac{1}{5})}{(1 - (-\frac{1}{5}))^3} = \frac{4/5}{(6/5)^3} = \frac{4/5}{216/125}$.
6. Calculate and simplify: $S = \frac{4}{5} \times \frac{125}{216} = \frac{4 \times 25}{216} = \frac{100}{216} = \frac{25}{54}$.