Let n$\in$N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms ${}^n{C_0},3.{}^n{C_1},5.{}^n{C_2},7.{}^n{C_3},.....$ is equal to 2100 . 101, then $2\left[ {{{n - 1} \over 2}} \right]$ is equal to _______________.
Answer (integer)
98
Solution
1. ${}^n{C_0} + 3.{}^n{C_1} + 5.{}^n{C_2} + ... + (2n + 1).{}^n{C_n}$<br><br>${T_r} = (2r + 1){}^n{C_r}$<br><br>$S = \sum {{T_r}}$<br><br>$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} }$<br><br>$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$<br><br>${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$<br><br>$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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