Let m, n N and gcd(2, n)=1. If 30 ( l 30 0 )+29 ( l 30 1 )+ +2 ( … - Vidyadip

MCQ

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

✓
$45$
B
$56$
C
$42$
D
$36$

✓

Answer

Correct option: A.

$45$

a
$30\left({ }^{30} C _{0}\right)+29\left({ }^{30} C _{1}\right)+\ldots+2\left({ }^{30} C _{28}\right)+1\left({ }^{30} C _{29}\right)$

$=30\left({ }^{30} C _{30}\right)+29\left({ }^{30} C _{29}\right)+\ldots \ldots+2\left({ }^{30} C _{2}\right)+1\left({ }^{30} C _{1}\right)$

$=\sum_{ r =1}^{30} r \left({ }^{30} C _{ r }\right)$

$=\sum_{ r =1}^{30} r \left(\frac{30}{ r }\right)\left({ }^{29} C _{ r -1}\right)$

$=30 \sum_{ r =1}^{30}{ }^{29} C _{ r -1}$

$=30\left({ }^{29} C _{0}+{ }^{29} C _{1}+{ }^{29} C _{2}+\ldots+{ }^{29} C _{29}\right)$

$=30\left(2^{29}\right)=15(2)^{30}= n (2)^{ m }$

$\therefore n =15, m =30$

$n + m =45$

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