Let $ m, n \in N $ and $ g c d(2, n)=1 . $ If $ 30\left(\begin{array}{c}30 \ 0\end{array}\right)+29\left(\begin{array}{c}30 \ 1\end{array}\right)+\ldots \ldots+2\left(\begin{array}{c}30 \ 28\end{array}\right)+1\left(\begin{array}{c}30 \ 29\end{array}\right)=n .2^{m} $, then $ n+m $ is equal to $ \left(\right. $ Here $ \left.\left(\begin{array}{l}n \ k\end{array}\right)={ }^{n} C_{k}\right) $

Question

Let $ m, n \in N $ and $ g c d(2, n)=1 . $ If $ 30\left(\begin{array}{c}30 \ 0\end{array}\right)+29\left(\begin{array}{c}30 \ 1\end{array}\right)+\ldots \ldots+2\left(\begin{array}{c}30 \ 28\end{array}\right)+1\left(\begin{array}{c}30 \ 29\end{array}\right)=n .2^{m} $, then $ n+m $ is equal to $ \left(\right. $ Here $ \left.\left(\begin{array}{l}n \ k\end{array}\right)={ }^{n} C_{k}\right) $,

MARKS App · Accepted Answer

30 C 0 30 + 29 C 1 30 + … + 2 C 28 30 + 1 C 29 30 = 30 C 30 30 + 29 C 29 30 + … … + 2 C 2 30 + 1 C 1 30 = ∑ r = 1 30 r C r 30 = ∑ r = 1 30 r 30 r C r - 1 29 = 30 ∑ r = 1 30 C r - 1 29 = 30 C 0 29 + C 1 29 + C 2 29 + … + C 29 29 = 30 2 29 = 15 2 30 = n 2 m ∴ n = 15 , m = 30 n + m = 45

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