Let C r denote the coefficient of x r in the binomial expansion of ( 1 + x ) n , n 0 r n . If P n = C 0 - C 1 + 2 2 3 ? C 2 - 2 3 4 ? C 3 + ? + ( - 2 ) n n + 1 ? C n , then the value of n = 1 25 1 P 2 n equals: [2026]

Question

Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1 + x)^{n}$ , $n \in ℕ$ $0 \leq r \leq n$ .

If $P_{n} = C_{0} - C_{1} + \frac{2^{2}}{3} C_{2} - \frac{2^{3}}{4} C_{3} + \dots + \frac{{(- 2)}^{n}}{n + 1} C_{n},$ then the value of $\sum_{n = 1}^{25} \frac{1}{P_{2 n}}$ equals: [2026]

Smart Achievers · Accepted Answer

(1)

$P_{n} = \sum_{r = 0}^{n} \frac{{}^{n}C_{r} {(- 2)}^{r}}{r + 1} = \sum_{r = 0}^{n} \frac{1}{n + 1} {}^{n + 1}C_{r + 1} {(- 2)}^{r}$

$= \frac{- 1}{2 (n + 1)} \sum_{r = 0}^{n} {}^{n + 1}C_{r + 1} {(- 2)}^{r + 1}$

$= \frac{- 1}{2 (n + 1)} [{(1 - 2)}^{n + 1} - 1]$

$P_{n} = \frac{1}{2 (n + 1)} [1 - {(- 1)}^{n + 1}]$

$P_{2 n} = \frac{1}{2 (2 n + 1)} [1 - {(- 1)}^{2 n + 1}]$

$P_{2 n} = \frac{1}{2 n + 1}$

$\sum_{n = 1}^{25} \frac{1}{P_{2 n}} = \sum_{n = 1}^{25} (2 n + 1)$

$= 3 + 5 + \dots + 51$

$= \frac{25}{2} (51 + 3)$

$= 25 \times 27 = 675$

Solution

1 675

2 525

3 650

4 580

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