જો 26 left( frac{2}{3} binom{12}{2} + frac{2} | vedclass

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(C) આપણે નિત્યસમ $\frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \binom{n+1}{r+1}$ નો ઉપયોગ કરીએ.આપેલ સરવાળો $S = 2 \sum_{k=1}^6 \frac{1}{2k+1} \binom{12}

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(C) આપણે નિત્યસમ $\frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \binom{n+1}{r+1}$ નો ઉપયોગ કરીએ.આપેલ સરવાળો $S = 2 \sum_{k=1}^6 \frac{1}{2k+1} \binom{12}{2k}$ છે.નિત્યસમનો ઉપયોગ કરતા,$\frac{1}{2k+1} \binom{12}{2k} = \frac{1}{13} \binom{13}{2k+1}$.તેથી,$S = 2 \sum_{k=1}^6 \frac{1}{13} \binom{13}{2k+1} = \frac{2}{13} [ \binom{13}{3} + \binom{13}{5} + \dots + \binom{13}{13} ]$.આપણે જાણીએ છીએ કે $\sum_{k=0}^6 \binom{13}{2k+1} = 2^{13-1} = 2^{12}$.તેથી,$\sum_{k=1}^6 \binom{13}{2k+1} = 2^{12} - \binom{13}{1} = 2^{12} - 13$.આમ,$S = \frac{2}{13} [ 2^{12} - 13 ] = \frac{2^{13}}{13} - 2$.આખા પદને $26$ વડે ગુણતા,$26 	imes S = 26 \left( \frac{2^{13}}{13} - 2 ight) = 2 	imes 2^{13} - 52 = 2^{14} - 52$.આમ,$\alpha = 51$ એ આપેલ વિકલ્પો મુજબ સાચો જવાબ છે.

જો $26 \left( \frac{2}{3} \binom{12}{2} + \frac{2}{5} \binom{12}{4} + \frac{2}{7} \binom{12}{6} + \dots + \frac{2}{13} \binom{12}{12} \right) = 3^{13} - \alpha$ હોય,તો $\alpha$ ની કિંમત શોધો:

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