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If the expansion of (1 + x)n, C0 + C1 + C2 + C3 + … Cn are coefficients different terms then find the value C0 + C2 + C4… . |
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Answer» (1 + x)n= nC0 1n + nC1 1n + nC1 1n-1 x1 nC2 1n-2 x2 + nC3 1n-3x3 +…. Putting x = 1 (1 + 1)n = nC0 + nC1 +nC2 + nC3 + …. Putting x = – 1 (1 – 1)n = nC0 – nC1 +nC2 – nC3 + …. Here nC0 + nC1 +nC2 + nC3 + …. = 2n … (i) “C0 – “Ci + ”C2 – “C3 + … = 0 … (ii) Adding equation (i) and (ii) 2[nC0 + nC2 + nC4 + …] = 2n ⇒ nC0 + nC2 + nC4 + … = 2n – 1 or C0 + C2 + C4 + … = 2n – 1 |
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