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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+..........^nC_nx^n$.....(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}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...