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

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Question

(b) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + ..... + {C_n}{x^n}$

${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + ..... +

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + ..... + {C_n}{x^n}$

${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + ..... + {( - 1)^n}{C_n}{x^n}$

$[{(1 + x)^n} - {(1 - x)^n}] = 2\,[{C_1}x + {C_3}{x^3} + {C_5}{x^5} + ...]$

$\frac{1}{2}[{(1 + x)^n} - {(1 - x)^n}] = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + .......$

Put $x = 2$, $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + .....\, = \frac{{{3^n} - {{( - 1)}^n}}}{2}$

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

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