sumlimits_{k = 0}^{10} {^{20}{C_k} = } | vedclass

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

Question

(A) Let $S = \sum\limits_{k = 0}^{10} {^{20}C_k} = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_{10}}$.We know that $\sum\limits_{k = 0}^{20} {^{20}C_k}

Vedclass · Accepted Answer

$2^{19} + \frac{1}{2} {^{20}C_{10}}$

Vedclass · Answer

(A) Let $S = \sum\limits_{k = 0}^{10} {^{20}C_k} = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_{10}}$.We know that $\sum\limits_{k = 0}^{20} {^{20}C_k} = 2^{20}$.Since ${^{n}C_r} = {^{n}C_{n-r}}$,we have ${^{20}C_0} = {^{20}C_{20}}$,${^{20}C_1} = {^{20}C_{19}}$,...,${^{20}C_9} = {^{20}C_{11}}$.Thus,$2^{20} = ({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}} + ({^{20}C_{11}} + \dots + {^{20}C_{20}})$.$2^{20} = 2({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}}$.Let $S = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9} + {^{20}C_{10}}$.Then $2S = 2({^{20}C_0} + \dots + {^{20}C_9}) + 2{^{20}C_{10}}$.Substituting the sum,$2S = (2^{20} - {^{20}C_{10}}) + 2{^{20}C_{10}} = 2^{20} + {^{20}C_{10}}$.Therefore,$S = 2^{19} + \frac{1}{2} {^{20}C_{10}}$.

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$

Similar Questions

In the expansion of $(1+a)^{m+n},$ prove that the coefficients of $a^{m}$ and $a^{n}$ are equal.

If $\sum_{r=0}^{20} {}^{20+r}C_r = \frac{p}{q} {}^{40}C_{20}$ and $GCD(p, q) = 1$,then $p^2 - q^2 =$

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $

$^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = ?$

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $,then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module