Question

(c) find the sum of each of the following series.
(i) $\sum_{n = 2}^{\infty}n(n - 1)x^{n}$, $|x| < 1$
(ii) $\sum_{n = 2}^{\infty}\frac{n^{2}-n}{3^{n}}$
(iii) $\sum_{n = 1}^{\infty}\frac{n^{2}}{2^{n}}$

Explanation:

Step1: Recall the geometric series formula

We know that the geometric series \(\sum_{n = 0}^{\infty}x^{n}=\frac{1}{1 - x}\), for \(|x|\lt1\). Differentiating both sides with respect to \(x\), we get \(\sum_{n = 1}^{\infty}nx^{n - 1}=\frac{1}{(1 - x)^{2}}\). Differentiating again, \(\sum_{n = 2}^{\infty}n(n - 1)x^{n - 2}=\frac{2}{(1 - x)^{3}}\).

Step2: Adjust the index for part (i)

Multiply the equation \(\sum_{n = 2}^{\infty}n(n - 1)x^{n - 2}=\frac{2}{(1 - x)^{3}}\) by \(x^{2}\). Then \(\sum_{n = 2}^{\infty}n(n - 1)x^{n}=\frac{2x^{2}}{(1 - x)^{3}}\) for \(|x|\lt1\).

Step3: Solve part (ii)

For \(\sum_{n = 2}^{\infty}\frac{n^{2}-n}{3^{n}}\), let \(x=\frac{1}{3}\). Using the result from part (i), \(\sum_{n = 2}^{\infty}n(n - 1)x^{n}=\frac{2x^{2}}{(1 - x)^{3}}\). Substituting \(x = \frac{1}{3}\), we have \(\sum_{n = 2}^{\infty}\frac{n(n - 1)}{3^{n}}=\frac{2\times(\frac{1}{3})^{2}}{(1-\frac{1}{3})^{3}}\).

$$
LATEXBLOCK0
$$

Step4: Solve part (iii)

We know that \(\sum_{n = 1}^{\infty}n^{2}x^{n}=x\sum_{n = 1}^{\infty}n^{2}x^{n - 1}\). First, \(\sum_{n = 1}^{\infty}nx^{n}=\frac{x}{(1 - x)^{2}}\) and \(\sum_{n = 1}^{\infty}n^{2}x^{n - 1}=\frac{1 + x}{(1 - x)^{3}}\). So \(\sum_{n = 1}^{\infty}n^{2}x^{n}=\frac{x(1 + x)}{(1 - x)^{3}}\). Let \(x=\frac{1}{2}\), then \(\sum_{n = 1}^{\infty}\frac{n^{2}}{2^{n}}=\frac{\frac{1}{2}(1+\frac{1}{2})}{(1-\frac{1}{2})^{3}}\).

$$
LATEXBLOCK1
$$

Answer:

(i) \(\frac{2x^{2}}{(1 - x)^{3}}\)
(ii) \(\frac{3}{4}\)
(iii) \(6\)