If {C_0}, {C_1}, {C_2}, ......., {C_n} are th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know the binomial expansions:$(1 + x)^n = {C_0} + {C_1}x + {C_2}x^2 + {C_3}x^3 + ..... + {C_n}x^n$$(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - {C_3

Vedclass · Accepted Answer

$\frac{{3^n - (-1)^n}}{2}$

Vedclass · Answer

(B) We know the binomial expansions:$(1 + x)^n = {C_0} + {C_1}x + {C_2}x^2 + {C_3}x^3 + ..... + {C_n}x^n$$(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - {C_3}x^3 + ..... + (-1)^n{C_n}x^n$Subtracting the two equations:$(1 + x)^n - (1 - x)^n = 2({C_1}x + {C_3}x^3 + {C_5}x^5 + .....)$Dividing by $2$:$\frac{(1 + x)^n - (1 - x)^n}{2} = {C_1}x + {C_3}x^3 + {C_5}x^5 + .....$To get the required series $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$,we substitute $x = 2$ in the expression ${C_1}x + {C_3}x^3 + {C_5}x^5 + .....$:$2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ..... = \frac{(1 + 2)^n - (1 - 2)^n}{2} = \frac{3^n - (-1)^n}{2}$

If ${C_0}, {C_1}, {C_2}, ......., {C_n}$ are the binomial coefficients,then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals

Similar Questions

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then $C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$ equals

The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + \dots + a_{2n}$ is equal to

$^nC_0 - \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 - \dots + (-1)^n \frac{^nC_n}{n+1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module