Let ^nC_r denote the binomial coefficient of x r in the expansion of (1 + x) n . If _k = 0^10 (2^2 + 3k) ^10C_k = .3^10 + .2^10, , R, then + is equal to ___________.

_k = 0^10 (2^2 + 3k)^10C_k = 4_k = 0^10 ^10C_k + 3_k = 0^10 k.^10C_k = 4(2^10) + 3_k = 0^10 k.10 k.^9C_k - 1 = 4(2^10) + 3.10(2^9) = 4(2^10) + 3.5.2^10 = 2^10(19) According to question, 19(2^10) = .3^10 + .2^10 = 0, = 19 + = 19

Let nCr denote the binomial coefficient of xr in the expansi: null (Mathematics)

\sum\limits_{k = 0}^{10} {({2^2} + 3k){}^{10}{C_k}}

= 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}}

= 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}}

4({2^{10}}) + 3.10({2^9})

= 4({2^{10}}) + {3.5.2^{10}}

= {2^{10}}(19)

According to question,

19({2^{10}}) = \alpha {.3^{10}} + \beta {.2^{10}}

\therefore

\alpha = 0,\beta = 19

\Rightarrow \alpha + \beta = 19

Explanation

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