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7.Binomial Theorem
easy
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
A
${2^{n - 1}}$
B
${2^{n - 1}}$
C
${2^n}$
D
${2^{n - 1}} - 1$
Solution
${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + …. + {C_n}{x^n}$
$x = 1$ 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: ${C_0} + {C_2} + {C_4} + …. = {2^{n – 1}}$.
Standard 11
Mathematics
