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Arithmetic Mean

Let 1+x221+...

Question

Let $(1 + x^{2})^{2} (1 + x)^{n} = \sum_{k = 0}^{n + 4} a_{k} x^{k}$ . If $a_{1}, a_{2}$ and $a_{3}$ are in arithmetic progression, then $n$ is

A

6

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B

4

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C

3

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D

2

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Solution

The correct options are
B $4$
C $2$
D $3$
L.H.S $= (1 + 2 x^{2} + x^{4})$ $\times (1 + c_{1} x + c_{2} x^{2} + c_{3} x^{3} + . . . . . .)$

R.H.S, $= a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . .$

Comparing the coefficients of $x, x^{2}, x^{3}, . . . . . . .$
$a_{1} = c_{1}, a_{2} = c_{2} + 2, a_{3} = c_{3} + 2 c_{1}$ .... (1)
Now $2 a_{2} = a_{1} + a_{3} (A . P .)$
$2 (^{n} C_{2} + 2) =^{n} C_{1} + (^{n} C_{3} + 2^{n} C_{1})$ ...By (1)
$\Rightarrow 2 \frac{n (n - 1)}{2} + 4 = 3 n + \frac{n (n - 1) (n - 2)}{6}$
$\Rightarrow n^{3} - 9 n^{2} + 26 n - 24 = 0$
Therefore, $(n - 2) (n^{2} - 7 n + 12) = 0$
$\Rightarrow (n - 2) (n - 3) (n - 4) = 0$
$\Rightarrow n = 2, 3, 4$

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