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EAMCET Engineering 2008 Solved Paper

Section: Mathematics

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Question : 89 of 160

Marks: +1, -0

If

(1+x+x^2+x^3)^5=\sum\limits_{k=0}^{15} a_k x^k

\sum\limits_{k=0}^{7} a_{2k}

128
256
512
1024

Solution:

\;\; \text{ Given, } \; (1+x+x^2+x^3)^5=\sum\limits_{k=0}^{15} a_k x^k

\Rightarrow [(1+x)+x(1+x)]^5=\sum\limits_{k=0}^{15} a_k x^k

\Rightarrow (1+x)^{10}=a_0 x^0+a_1 x+a_2 x^2+\ldots+a_{15} x^{15}

\Rightarrow \;\; {}^{10}C_0+{}^{10}C_1 x+{}^{10}C_2 x^2+\ldots+{}^{10}C_{10} x^{10}

=a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots+a_{15} x^{15}

On equating the coefficient of constant and even powers of

x

, we get

a_0\;={}^{10}C_0, a_2={}^{10}C_2

a_4\;={}^{10}C_4, \ldots, a_{10}={}^{10}C_{10}

a_{12}\;=a_{14}=0

\therefore \sum\limits_{k=0}^{7} a_{2k} = {}^{10}C_0+{}^{10}C_2+{}^{10}C_4+{}^{10}C_6

\; +{}^{10}C_8+{}^{10}C_{10}+0+0

\; =2^{10-1}=2^9

\; =512

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