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

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Question

${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$

$x = 1$ रखने पर,

$ \Rightarrow {2^n} = {C_0} + {C_1} + {C_2} + ..... + {C_n}$...

Vedclass · Accepted Answer

${2^{n - 1}}$

Vedclass · Answer

${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$

$x = 1$ रखने पर,

$ \Rightarrow {2^n} = {C_0} + {C_1} + {C_2} + ..... + {C_n}$.....(i)

या ${C_1} + {C_2} + {C_3} + .... + {C_n} = {2^n} - 1$

पुन: $x = -1$ रखने पर,

  $0 = {C_0} - {C_1} + {C_2} - {C_3} + ....$

या ${C_0} + {C_2} + {C_4} + ....$$ = {C_1} + {C_3} + {C_5} + ....$अर्थात् $A = B$

(i) से $A + B  = 2n$  या $A = {2^{n - 1}} = B$

अत: ${C_0} + {C_2} + {C_4} + .... = {2^{n - 1}}$.

यदि ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}$, तब ${C_0} + {C_2} + {C_4} + {C_6} + .....$ का मान होगा

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