If {(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} | vedclass

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Question

(a) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$

Putting $x = 1$, we get

$ \Rightarrow {2^n} = {C_0} + {C_1} + {C_2} + .....

Vedclass · Accepted Answer

${2^{n - 1}}$

Vedclass · Answer

(a) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$

Putting $x = 1$, we get

$ \Rightarrow {2^n} = {C_0} + {C_1} + {C_2} + ..... + {C_n}$.....(i)

or ${C_1} + {C_2} + {C_3} + .... + {C_n} = {2^n} - 1$      $[\,{C_0} = {\,^n}{C_0} = 1]$

Again, putting $x = -1$, we get

$0 = {C_0} - {C_1} + {C_2} - {C_3} + ....$

or ${C_0} + {C_2} + {C_4} + ....$$ = {C_1} + {C_3} + {C_5} + ....$ $ i.e. A = B$

Also from  $(i), A + B =2^n$ or $A = {2^{n - 1}} = B$

Hence, ${C_0} + {C_2} + {C_4} + .... = {2^{n - 1}}$.

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}$, then the value of ${C_0} + {C_2} + {C_4} + {C_6} + .....$ is

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