The value of sumlimits_{r = 1}^{19} {frac{{{} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $S = \sum\limits_{r = 1}^{19} \frac{{}^{20}C_{r+1}(-1)^r}{2^{2r+1}}$.We can rewrite the term as $\frac{1}{2} \sum\limits_{r=1}^{19} {}^{20}C_{

Vedclass · Accepted Answer

$-2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} + 4} \right)$

Vedclass · Answer

(B) Let $S = \sum\limits_{r = 1}^{19} \frac{{}^{20}C_{r+1}(-1)^r}{2^{2r+1}}$.We can rewrite the term as $\frac{1}{2} \sum\limits_{r=1}^{19} {}^{20}C_{r+1} \left(-\frac{1}{4}\right)^r$.Let $k = r+1$,then as $r$ goes from $1$ to $19$,$k$ goes from $2$ to $20$.$S = \frac{1}{2} \sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^{k-1} = \frac{1}{2} \sum\limits_{k=2}^{20} {}^{20}C_k (-4) \left(-\frac{1}{4}\right)^k = -2 \sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^k$.Using the binomial expansion $(1+x)^n = \sum\limits_{k=0}^n {}^{n}C_k x^k$,we have $\sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^k = (1 - \frac{1}{4})^{20} - {}^{20}C_0 - {}^{20}C_1(-\frac{1}{4})$.$= (\frac{3}{4})^{20} - 1 - 20(-\frac{1}{4}) = (\frac{3}{4})^{20} - 1 + 5 = (\frac{3}{4})^{20} + 4$.Therefore,$S = -2 \left( (\frac{3}{4})^{20} + 4 \right)$.

The value of $\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}{(-1)}^r}}{{{2^{2r + 1}}}}}$ is

Similar Questions

If $f(x) = x^n$,then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!}$ is

The sum of the coefficients of the terms of degree $m$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:

If $(3+\sqrt{2})^6-(3-\sqrt{2})^6=a+b \sqrt{2}$,then $a+b=$

Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $(x+\sqrt{x^3-1})^5+(x-\sqrt{x^3-1})^5, x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$,then $u+v$ equals:

For $|x| < \frac{1}{5}$,the coefficient of $x^3$ in the expansion of $\frac{1}{(1-5 x)(1-4 x)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module