Let m, n in N and operatorname{gcd}(2, n)=1. | vedclass

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Question

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

Vedclass · Accepted Answer

$45$

Vedclass · Answer

$30\left({ }^{30} C _{0}ight)+29\left({ }^{30} C _{1}ight)+\ldots+2\left({ }^{30} C _{28}ight)+1\left({ }^{30} C _{29}ight)$

$=30\left({ }^{30} C _{30}ight)+29\left({ }^{30} C _{29}ight)+\ldots \ldots+2\left({ }^{30} C _{2}ight)+1\left({ }^{30} C _{1}ight)$

$=\sum_{ r =1}^{30} r \left({ }^{30} C _{ r }ight)$

$=\sum_{ r =1}^{30} r \left(\frac{30}{ r }ight)\left({ }^{29} C _{ r -1}ight)$

$=30 \sum_{ r =1}^{30}{ }^{29} C _{ r -1}$

$=30\left({ }^{29} C _{0}+{ }^{29} C _{1}+{ }^{29} C _{2}+\ldots+{ }^{29} C _{29}ight)$

$=30\left(2^{29}ight)=15(2)^{30}= n (2)^{ m }$

$	herefore n =15, m =30$

$n + m =45$

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