Let ${(1 + x^{2})}^{2} {(1 + x)}^{n} = A_{0} + A_{1} x + A_{2} x^{2} + . . . .$ If $A_{0}, A_{1}, A_{2}$ are in A.P, then the value of $n$

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Solution

The correct option is A $2$
According to the question:
$\begin{matrix} (1 + x^{2})^{2} (1 + x)^{n} = A_{0} + A_{1} x + A_{2} x^{2} - - - - - - - (1) p u t x = 0 t h e n (1) (1) = A_{o} ∴ A_{o} = 1 N o w, d i f f e r e n t i a t i n g e q u : (1) 2 (1 + x^{2}) (2 x) (1 + x)^{n} + n (1 + n)^{n - 1} (1 + x^{2})^{2} = A_{1} + 2 A_{2} \times x A g a i n, p u t x = 0 n = A_{1} o r A_{1} = n A g a i n, d i f f e r e n t i a t n g w . r . t, 2.2 x (2 x) (1 + x)^{n} + 4 (1 + x^{2}) (1 + x)^{n} + 2 n (1 + x^{2}) (2 x) (1 + n)^{n - 1} + n (n - 1) (1 + x)^{n - 2} (1 + x^{2})^{2} + n 2 (1 + x^{n - 1}) (1 + x^{2}) 2 x = 2 A_{2} p u t x = 0 \Rightarrow 4 + n (n - 1) = 2 A_{2} A_{2} = \frac{4 + n^{2} - n}{2} A_{0}, A_{1}, A_{2}, - - \to A . P W e c a n s a y, \Rightarrow 1, n, \frac{n^{2} - n + 4}{2} \Rightarrow n - 1 = \frac{n^{2} - n + 4}{2} - n \Rightarrow 2 n - 1 = \frac{n^{2} - n + 4}{2} \Rightarrow 4 n - 2 = n^{2} - n + 4 \Rightarrow n^{2} - 5 n + 6 = 0 \Rightarrow n^{2} - 3 n - 2 n + 6 = 0 \Rightarrow n (n - 3) - 2 (n - 3) = 0 \Rightarrow (n - 3) (n - 2) ∴ n = 2, 3 \end{matrix}$

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