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In the expansion of (1 + x)n, what is the sum of even binomial coefficients?

In the expansion of (1 + x)n, what is the sum of even binomial coefficients? Correct Answer 2<sup>n - 1</sup>

Concept:

(1 + x)n = nC0 x0+ nC1 x1 + nC2 x2 + …. + nCn xn

Calculation:

To find: Sum of even binomial coefficients

That is,nC0 + nC2 + nC4 + nC6 + nC8 + …. 

As we know, (1 + x)n = nC0 x0+ nC1 x1 + nC2 x2 + …. + nCn xn

Put x = 1, we get

⇒ (1 + 1)n = nC0 10+ nC1 11 + nC2 12 + …. + nCn 1n

⇒ nC0 10+ nC1 11 + nC2 12 + …. + nCn 1n = 2n

∴ nC0 + nC1 + nC2 + …. + nCn = 2n                           .... (1)

Put x = -1, we get

⇒ (1 - 1)n = nC0 (-1)0+ nC1 (-1)1 + nC2 (-1)2 + …. + nCn (-1)n

∴ nC0 - nC1 + nC2 - …. +(-1)n nCn = 0                       .... (2)

Adding eqaution (1) + (2), we get

2 × (nC0 + nC2 + nC4 + nC6 + nC8 + ….) = 2n + 0

∴ nC0 + nC2 + nC4 + nC6 + nC8 + ….= 2n - 1

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