Question
Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1 + x)^{n},n \in \mathbb{N},0 \leq r \leq n$.
If $P_{n} = C_{0} - C_{1} + \frac{2^{2}}{3}C_{2} - \frac{2^{3}}{4}C_{3} + \ldots.. + \frac{( - 2)^{n}}{n + 1}C_{n}$, then the value of $\sum_{n = 1}^{25}\mspace{2mu}\frac{1}{P_{2n}}$ equals.
Options
Solution
$P_{n} = \sum_{r = 0}^{n}\mspace{2mu}\frac{\ ^{n}C_{r}( - 2)^{r}}{r + 1} = \sum_{r = 0}^{n}\mspace{2mu}\frac{1}{(n + 1)}\ ^{n + 1}C_{r + 1}( - 2)^{r}$
$${= \frac{- 1}{2(n + 1)}\sum_{r = 0}^{n}\mspace{2mu}\ ^{n + 1}C_{r + 1}( - 2)^{r + 1} }{= \frac{- 1}{2(n + 1)}\left\lbrack (1 - 2)^{n + 1} - 1 \right\rbrack }{P_{n} = \frac{1}{2(n + 1)}\left\lbrack 1 - ( - 1)^{n + 1} \right\rbrack}$$
$$\begin{matrix} & P_{2n} = \frac{1}{2(2n + 1)}\left\lbrack 1 - ( - 1)^{2n + 1} \right\rbrack \\ & P_{2n} = \frac{1}{2n + 1} \\ & \ \sum_{n = 1}^{25}\mspace{2mu}\mspace{2mu}\frac{1}{P_{2n}} = \sum_{n = 1}^{25}\mspace{2mu}\mspace{2mu}(2n + 1) \\ & \ = 3 + 5 + \ldots + 51 \\ & \ = \frac{25}{2}\lbrack 51 + 3\rbrack \\ & \ = 25 \times 27 = 675 \end{matrix}$$
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