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UPSC NDA II 2015 Mathematics Paper

Question Numbers: 89-91

For the next three (03) items that follow:

Consider the expansion of

(1 + x)^{2n + 1}

Question : 90 of 120

Marks: +1, -0

The average of the coefficients of the two middle terms in the expansion is

Solution:

Concept:

In a binomial expansion

(1+x)^{2n+1}

, the power is odd, so there are two middle terms. Their coefficients are added and averaged using binomial identities.

Explanation:

For

(1+x)^{2n+1}

, total terms =

2n+2

. Since

2n+1

is odd, the two middle terms are the

(n+1)

th and

(n+2)

th terms.

The coefficient of the

r

th term is

^{2n+1}C_{r-1}

. Hence the two coefficients are

^{2n+1}C_n

and

^{2n+1}C_{n+1}

Average of the coefficients =

\frac{^{2n+1}C_n + ^{2n+1}C_{n+1}}{2}

Using the identity

^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}

, we get:

^{2n+1}C_n + ^{2n+1}C_{n+1} = ^{2n+2}C_{n+1}

Thus average =

\frac{^{2n+2}C_{n+1}}{2}

Now apply

{}^nC_r = \frac{n}{r} \, {}^{n-1}C_{r-1}

^{2n+2}C_{n+1} = \frac{2n+2}{n+1} \cdot ^{2n+1}C_n = 2 \cdot ^{2n+1}C_n

Therefore average =

\frac{2 \cdot ^{2n+1}C_n}{2} = ^{2n+1}C_n

Answer:

Option B:

^{2n+1}C_n

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