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The sum of coefficients of odd powers of $x$ in the expansion $(1+x)^{50}$ is ?

$\begin{array}{1 1} 0 \\ 2^{49} \\ 2^{50} \\ 2^{51} \end{array}$

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$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+..........^nC_nx^n$.....(i)
$(1-x)^n=^nC_0-^nC_1x+^nC_2x^2-..........^nC_nx^n$....(ii)
$=(^nC_0+^nC_2x^2+^nC_4+......)$ - $(^nC_1x+^nC_3x^3+^nC_5x^5+........)$
Take $x=1,\:\:and\:\:n=50$ in $(i)$
$2^{50}=^nC_0+^nC_1+^nC_2+.........^nC_n$
$i.e.,$$ 2^{50}=$ (sum of coeff. of odd powers of $x$)+(sum of coeff. of even powers of $x$
Similarly by taking $x=1,n=50$ in $(ii)$
$0$=(Sum of coeff. of odd powers)-(sum of coeff. of even powers)
$\Rightarrow\:$sum of coeff. of odd powers = sum of coeff. of even powers.
$\therefore 2^{50}=2$(sum of coeff. of odd powers)
$\Rightarrow\:$Sum of coeff. of odd powers =$ 2^{49}$

 

answered Sep 24, 2013 by rvidyagovindarajan_1
edited Dec 24, 2013 by meenakshi.p
 

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