If sum_{r=0}^{20} {}^{20+r}C_r = frac{p}{q} { | vedclass

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

Question

(C) We know the identity ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.Given the sum $S = \sum_{r=0}^{20} {}^{20+r}C_r = {}^{20}C_0 + {}^{21}C_1 + {}^

Vedclass · Accepted Answer

$1240$

Vedclass · Answer

(C) We know the identity ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.Given the sum $S = \sum_{r=0}^{20} {}^{20+r}C_r = {}^{20}C_0 + {}^{21}C_1 + {}^{22}C_2 + \dots + {}^{40}C_{20}$.Since ${}^{20}C_0 = 1 = {}^{21}C_0$,we can write $S = {}^{21}C_0 + {}^{21}C_1 + {}^{22}C_2 + \dots + {}^{40}C_{20}$.Using the identity ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ repeatedly:${}^{21}C_0 + {}^{21}C_1 = {}^{22}C_1$.Then ${}^{22}C_1 + {}^{22}C_2 = {}^{23}C_2$.Continuing this process,the sum telescopes to ${}^{40}C_{20} + {}^{40}C_{21} = {}^{41}C_{21}$.Now,${}^{41}C_{21} = \frac{41}{21} 	imes {}^{40}C_{20}$.Thus,$\frac{p}{q} = \frac{41}{21}$,which gives $p = 41$ and $q = 21$.Since $GCD(41, 21) = 1$,we calculate $p^2 - q^2 = 41^2 - 21^2 = (41 - 21)(41 + 21) = 20 	imes 62 = 1240$.

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

Similar Questions

The sum of the last $30$ coefficients in the expansion of $(1+x)^{59}$, when expanded in ascending powers of $x$, is:

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

$\sum_{r=1}^{15} r^2 \left( \frac{{}^{15}C_r}{{}^{15}C_{r-1}} \right) = $

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

If $a$ and $d$ are two complex numbers,then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module