If binom{40}{0} + binom{41}{1} + binom{42}{2} | vedclass

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(A) We use the identity $\binom{n}{r} + \binom{n+1}{r+1} = \binom{n+2}{r+1} - \binom{n}{r+1}$ or the property $\binom{n}{r} = \binom{n-1}{r} + \binom{

Vedclass · Accepted Answer

$102$

Vedclass · Answer

(A) We use the identity $\binom{n}{r} + \binom{n+1}{r+1} = \binom{n+2}{r+1} - \binom{n}{r+1}$ or the property $\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$.Let $S = \sum_{k=0}^{20} \binom{40+k}{k}$.Using the identity $\sum_{k=0}^{r} \binom{n+k}{k} = \binom{n+r+1}{r}$,we have:$S = \binom{40+20+1}{20} = \binom{61}{20}$.We are given $S = \frac{m}{n} \binom{60}{20}$.So,$\binom{61}{20} = \frac{m}{n} \binom{60}{20}$.Using $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$,we get $\binom{61}{20} = \frac{61}{20+41} \dots$ actually $\binom{61}{20} = \frac{61}{61-20} \binom{60}{20} = \frac{61}{41} \binom{60}{20}$.Thus,$m = 61$ and $n = 41$.Since $m$ and $n$ are coprime,$m+n = 61 + 41 = 102$.

If $\binom{40}{0} + \binom{41}{1} + \binom{42}{2} + \dots + \binom{60}{20} = \frac{m}{n} \binom{60}{20}$,where $m$ and $n$ are coprime,then $m+n$ is equal to

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