Σlimits_{k = 0}^{10} {^{20}{Cₖ} = }

English

Gujarati

Hindi

7.Binomial Theorem

medium

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

A

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

B

${2^{19}}$

C

$^{20}{C_{10}}$

D

None of these

Solution

(a) $\sum\limits_{K = 0}^{10} {^{20}{C_k}} $ i.e., $^{20}{C_0} + {\,^{20}}{C_1} + …… + {\,^{20}}{C_{10}}$

We know that, ${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}{x^1} + {\,^n}{C_2}{x^2} + …. + {\,^n}{C_n}.{x^n}$

Put $x = 1$; ${2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ….. + {\,^n}{C_n}$

Put $n = 20$; ${2^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + …… + {\,^{20}}{C_{20}}$

${2^{20}} + \,{\,^{20}}{C_{10}} = 2\,[{\,^{20}}{C_0} + {\,^{20}}{C_1} + …… + {\,^{20}}{C_{10}}]$

${[^{20}}{C_0} + {\,^{20}}{C_1} + …… + {\,^{20}}{C_{10}}] = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$

$\sum\limits_{k = 0}^{10} {^{20}{C_k}} = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$.

Standard 11

Mathematics

Share

Similar Questions

If $^{20}{C_1} + \left( {{2^2}} \right){\,^{20}}{C_3} + \left( {{3^2}} \right){\,^{20}}{C_3} + \left( {{2^2}} \right) + ….. + \left( {{{20}^2}} \right){\,^{20}}{C_{20}} = A\left( {{2^\beta }} \right)$, then the ordered pair $(A, \beta )$ is equal to

hard

(JEE MAIN-2019)

For integers $n$ and $r$, let $\left(\begin{array}{l} n \\ r \end{array}\right)=\left\{\begin{array}{ll}{ }^{n} C _{ r }, & \text { if } n \geq r \geq 0 \\ 0, & \text { otherwise }\end{array}\right.$

The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\left(\begin{array}{c}10 \\ i\end{array}\right)\left(\begin{array}{c}15 \\ k-i\end{array}\right)+\sum_{i=0}^{k+1}\left(\begin{array}{c}12 \\ i\end{array}\right)\left(\begin{array}{c}13 \\ k+1-i\end{array}\right)$ exists, is equal to …… .

medium

(JEE MAIN-2021)

The sum of coefficients in ${(1 + x – 3{x^2})^{2134}}$ is

easy

Given $(1 – 2x + 5x^2 – 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + ….$ and that $a_1^2\,= 2a_2$ then the value of $n$ is

normal

The coefficient of $x^{91}$ in the series $^{100}{C_1}\,{2^8}.\,{\left( {1\, – \,x} \right)^{99}}\, + {\,^{100}}{C_2}\,{2^7}.\,{\left( {1\, – \,x} \right)^{98}}\, + {\,^{100}}{C_3}\,{2^6}.\,{\left( {1\, – \,x} \right)^{97}}\, + \,….\, + {\,^{100}}{C_9}\,{\left( {1\, – \,x} \right)^{91}}$ is equal to –

normal