- Identify the series as derivatives of the geometric series: \(\sum_{n=0}^{\infty} x^n = (1-x)^{-1}\).
- Use the formula \(S_1(x) = \sum_{n=2}^{\infty} n(n-1) x^n = \frac{2x^2}{(1-x)^3}\) for part (i).
- Substitute \(x = 1/4\) into the formula for part (ii) to get \(\frac{8}{27}\).
- Evaluate part (iii) by splitting the numerator \(n^2 = n(n-1) + n\) and summing the individual components at \(x = 1/2\), resulting in \(6\).
- Final answers: (i) \(\frac{2x^2}{(1-x)^3}\), (ii) \(\boxed{\frac{8}{27}}\), (iii) \(\boxed{6}\).
Explanation
- Deriving the Series Summation Formula (i)
To find the sum of the power series \(\sum_{n=2}^{\infty} n(n-1) x^n\), we begin with the standard geometric series: \(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) for \(|x|<1\). Taking the first derivative with respect to \(x\) gives \(\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}\). Differentiating again with respect to \(x\) yields \(\sum_{n=2}^{\infty} n(n-1)x^{n-2} = \frac{2}{(1-x)^3}\). Multiplying this series by \(x^2\) gives the desired result for (i): \(\sum_{n=2}^{\infty} n(n-1) x^n = \frac{2x^2}{(1-x)^3}\).
- Summation of Series (ii)
For series (ii), \(\sum_{n=2}^{\infty} \frac{n(n-1)}{4^n}\), we use the formula derived in step 1, \(S_1(x) = \frac{2x^2}{(1-x)^3}\), with \(x = \frac{1}{4}\). Substituting this value:
\[S_2 = \frac{2(\frac{1}{4})^2}{(1 - \frac{1}{4})^3} = \frac{2(\frac{1}{16})}{(\frac{3}{4})^3} = \frac{\frac{1}{8}}{\frac{27}{64}} = \frac{1}{8} \times \frac{64}{27} = \frac{8}{27}\]
Thus, the sum of the series is \(\frac{8}{27}\).
- Summation of Series (iii)
For series (iii), \(\sum_{n=1}^{\infty} \frac{n^2}{2^n}\), we use the identity \(n^2 = n(n-1) + n\). We can split the sum into two parts:
\[\sum_{n=1}^{\infty} \frac{n^2}{2^n} = \sum_{n=1}^{\infty} \frac{n(n-1)}{2^n} + \sum_{n=1}^{\infty} \frac{n}{2^n}\]
The first part \(\sum_{n=1}^{\infty} \frac{n(n-1)}{2^n}\) corresponds to \(S_1(1/2) = \frac{2(1/2)^2}{(1-1/2)^3} = \frac{2(1/4)}{(1/2)^3} = \frac{1/2}{1/8} = 4\).
The second part \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) corresponds to \(\frac{x}{(1-x)^2}\) evaluated at \(x = 1/2\), which is \(\frac{1/2}{(1-1/2)^2} = \frac{1/2}{1/4} = 2\).
Adding these results gives \(4 + 2 = 6\). The sum of the series is \(6\).
- Final Results
In summary, the sums of the series are:
(i) \(\frac{2x^2}{(1-x)^3}\)
(ii) \(\frac{8}{27}\)
(iii) \(6\)
All sums are determined by utilizing the geometric series and its derivatives.
Examples
Understanding how to sum infinite series is essential in fields like physics and engineering, particularly when analyzing signal processing or stability of systems. For example, in vibration analysis, the response of a system might be expressed as an infinite sum. Being able to compute these sums accurately allows engineers to predict the long-term behavior of structures under periodic loads, ensuring that designs are both safe and efficient. Just as we used derivatives of the geometric series to simplify these complex summations, similar techniques are used to solve differential equations that describe physical phenomena.