Let m, n$\in$N and gcd (2, n) = 1. If $$30\left( {\matrix{
{30} \cr
0 \cr
} } \right) + 29\left( {\matrix{
{30} \cr
1 \cr
} } \right) + ...... + 2\left( {\matrix{
{30} \cr
{28} \cr
} } \right) + 1\left( {\matrix{
{30} \cr
{29} \cr
} } \right) = n{.2^m}$$, then n + m is equal to __________.
(Here $\left( {\matrix{
n \cr
k \cr
} } \right) = {}^n{C_k}$)
Answer (integer)
45
Solution
$$30({}^{30}{C_0}) + 29({}^{30}{C_1}) + .... + 2({}^{30}{C_{28}}) + 1({}^{30}{C_{29}})$$<br><br>$$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$$<br><br>$= \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})}$<br><br>$$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({}^{29}{C_{r - 1}}} )$$<br><br>$= 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}}$<br><br>$= 30({}^{29}{C_0} + {}^{29}{C_1} + {}^{29}{C_2} + ..... + {}^{29}{C_{29}})$<br><br>$= 30({2^{29}}) = 15{(2)^{30}} = n{(2)^m}$<br><br>$\therefore$ n = 15, m = 30<br><br>$\Rightarrow$ n + m = 45
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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