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

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(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

$1240$

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(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 $p^2 - q^2 =$

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