Let m, n and gcd (2, n) = 1. If 30 $300 + 29 301 + + 2 3028 + 1 3029 = n 2^m \\\_\_\_\_\_\_\_\_\_\_$....

30(^30C_0) + 29(^30C_1) + .... + 2(^30C_28) + 1(^30C_29) = 30(^30C_30) + 29(^30C_29) + ...... + 2(^30C_2) + 1(^30C_1) = _r = 1^30 r(^30C_r) = _r = 1^30 r( 30 r )(^29C_r - 1 ) = 30_r = 1^30 ^29C_r - 1 = 30(^29C_0 + ^29C_1 + ^29C_2 + ..... + ^29C_29) = 30(2^29) = 15(2)^30 = n(2)^m n = 15, m = 30 n + m...

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Let m nN and gcd 2 n 1 If 30 30 0 29 30 1 2 30 28 1 3: null (Mathematics) | DivineJEE

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Mathematics

Exam

Mathematics

Let m, n

\in \mathbb{N}

and gcd (2, n) = 1. If 30

\binom{30}{0} + 29 \binom{30}{1} + \dots + 2 \binom{30}{28} + 1 \binom{30}{29} = n \cdot 2^m \\

then n + m is equal to

\_\_\_\_\_\_\_\_\_\_

\left( \text{Here } \binom{n}{k} = {}^nC_k \right)

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