સરવાળો શોધો: left( binom{21}{1} - binom{10}{1 | vedclass

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Question

(A) આપેલ પદાવલિ $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$ છે.આપણે જાણીએ છીએ કે $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.$\su

Vedclass · Accepted Answer

$2^{20} - 2^{10}$

Vedclass · Answer

(A) આપેલ પદાવલિ $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$ છે.આપણે જાણીએ છીએ કે $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.$\sum_{r=0}^{10} \binom{21}{r} = 2^{20}$ થાય.તેથી,$\sum_{r=1}^{10} \binom{21}{r} = 2^{20} - \binom{21}{0} = 2^{20} - 1$.વળી,$\sum_{r=1}^{10} \binom{10}{r} = 2^{10} - 1$.કિંમતો મૂકતા:$S = (2^{20} - 1) - (2^{10} - 1) = 2^{20} - 2^{10}$.

સરવાળો શોધો: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

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