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Question: If (1 + x + x<sup>2</sup>)<sup>n</sup> = \(\sum_{r = 0}^{2n}{a_{r}x^{r}}\), then \(\sum_{r = 0}^{n}{...

If (1 + x + x2)n = ∑r=02narxr\sum_{r = 0}^{2n}{a_{r}x^{r}}, then ∑r=0n(−1)r\sum_{r = 0}^{n}{( - 1)^{r}}arnCr = nCn/3, if n is

A

3k + 1

B

3k + 2

C

3k

D

None of these

Answer

3k

Explanation

Solution

We have,

(1 + x + x2)n = a0 + a1x + a2x2 + ….. + a2nx2n …(1)

And (x – 1)n = nC0 xn – nC1 xn–1 + nC2 xn–2 –….. + (–1)n nCn xn

…(2)

Multiplying (1) and (2), we get

(1 + x + x2)n (x – 1)n

= (a0 + a1x + a2x2 + ….. + a2n x2n) × {nC0xn – nC1 xn–1

+ nC2 xn–2 –….+(–1)n nCn}

Ž (x3 – 1)n

= (a0 + a1x + a2x2 + …. + a2n x2n) × {nC0xn – nC1 xn–1 +….

+(–1)n nCn} ….(3)

Now, coefficient of xn of RHS of (3)

= a0 nC0 – a1nC1 + a2 nC2 – ….. + (–1)n an nCn

LHS of (3)

= (x3 –1)n

= (–1)n (1 – x3)n

= (–1)n ∑r=0nnCr\sum_{r = 0}^{n}{nC_{r}} (– x3)r

= (–1)n ∑r=0n(−1)r\sum_{r = 0}^{n}{( - 1)^{r}}nCr x3r ….(4)

Clearly, if n is not a multiple of 3, then xn does not occur in (4)

\ (Coefficient of xn in LHS) = 0,

When n is not a multiple of 3

If n is a multiple of 3 i.e. if n = 3m, then

= (–1)3m (–1)m 3mCm

= 3mCm = nCn/3 [Q n = 3m]

Thus, equation the coefficients of xn on both sides, we get

a0 nC0 – a1 nC1 + a2 nC2 – a3 nC3 + ….. + (–1)n an nCn

=

Hence (3) is correct answer.