Question ID: #772
Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in\mathbb{N}, 0 \le r \le n$.
If $P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3}C_{2}-\frac{2^{3}}{4}C_{3}+….+\frac{(-2)^{n}}{n+1}C_{n}$, then the value of $\sum_{n=1}^{25}\frac{1}{P_{2n}}$ equals:
- (1) 580
- (2) 525
- (3) 650
- (4) 675
Solution:
**Simplify $P_n$**
$$ P_n = \sum_{r=0}^{n} C_r \frac{(-2)^r}{r+1} \cdot \frac{1}{(-1)^0} \text{ (Adjusting signs)} $$
Actually, let’s write the general term correctly from the expansion:
$$ P_n = \sum_{r=0}^n \frac{(-1)^r 2^r}{r+1} C_r = \sum_{r=0}^n \frac{(-2)^r}{r+1} {}^nC_r $$
Using the identity $\frac{{}^nC_r}{r+1} = \frac{1}{n+1} {}^{n+1}C_{r+1}$:
$$ P_n = \frac{1}{n+1} \sum_{r=0}^n (-2)^r {}^{n+1}C_{r+1} $$
Let $k = r+1$. When $r=0, k=1$; when $r=n, k=n+1$.
$$ P_n = \frac{1}{n+1} \sum_{k=1}^{n+1} (-2)^{k-1} {}^{n+1}C_k $$
$$ P_n = \frac{1}{n+1} \cdot \frac{1}{-2} \sum_{k=1}^{n+1} (-2)^k {}^{n+1}C_k $$
Recall expansion $(1+x)^m = \sum C_k x^k$. Here $x=-2, m=n+1$.
$$ \sum_{k=0}^{n+1} {}^{n+1}C_k (-2)^k = (1-2)^{n+1} = (-1)^{n+1} $$
$$ {}^{n+1}C_0 + \sum_{k=1}^{n+1} {}^{n+1}C_k (-2)^k = (-1)^{n+1} $$
$$ 1 + \sum_{k=1}^{n+1} (-2)^k {}^{n+1}C_k = (-1)^{n+1} $$
$$ \sum_{k=1}^{n+1} (-2)^k {}^{n+1}C_k = (-1)^{n+1} – 1 $$
Substitute back into $P_n$:
$$ P_n = \frac{1}{-2(n+1)} [ (-1)^{n+1} – 1 ] = \frac{1}{2(n+1)} [ 1 – (-1)^{n+1} ] $$
**Find $P_{2n}$**
Replace $n$ with $2n$:
$$ P_{2n} = \frac{1}{2(2n+1)} [ 1 – (-1)^{2n+1} ] $$
Since $2n+1$ is odd, $(-1)^{2n+1} = -1$.
$$ P_{2n} = \frac{1}{2(2n+1)} [ 1 – (-1) ] = \frac{2}{2(2n+1)} = \frac{1}{2n+1} $$
**Calculate the Sum**
$$ \sum_{n=1}^{25} \frac{1}{P_{2n}} = \sum_{n=1}^{25} (2n+1) $$
This is an Arithmetic Progression: $3, 5, 7, \dots$ with 25 terms.
First term $a = 3$, Last term $l = 2(25)+1 = 51$.
$$ Sum = \frac{N}{2}(a+l) = \frac{25}{2}(3+51) = \frac{25}{2}(54) $$
$$ Sum = 25 \times 27 = 675 $$
Ans. (4)
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