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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

Question : 88 of 160

Marks: +1, -0

(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}

\sum\limits_{r=1}^{15} r \frac{a_r}{a_r-1}

Solution:

Given that

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

\Rightarrow {}^{15}C_0+{}^{15}C_1 x+{}^{15}C_2 x^2+\ldots+{}^{15}C_{15} x^{15}

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

Equating the coefficient of various powers of

x

, we get

\;a_0={}^{15}C_0, a_1={}^{15}C_1, a_2={}^{15}C_2, \ldots, a_{15}={}^{15}C_{15}

\therefore \;\sum\limits_{r=1}^{15} r \frac{a_r}{a_{r-1}}=\sum\limits_{r=1}^{15} r \frac{{}^{15}C_r}{{}^{15}C_{r-1}}

=\sum\limits_{r=1}^{15} r \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(15-r+1)!}}

=\;\sum\limits_{r=1}^{15} \frac{(r-1)!(15-r+1)!}{r!(15-r)!}

=\;\sum\limits_{r=1}^{15} 15-r+1=15+14+13+\ldots+2+1

=\;\frac{15(15+1)}{2}=120

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