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Sum of N Terms of an AP

Let 1+x22 1...

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 A.P, find sum of all possible values of $n$ ?

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Solution

$(1 + x^{2})^{2} (1 + x)^{n}$
$= (1 + 2 x^{2} + x^{4} (1 + n x +^{n} C_{2} x^{2} . . .)$
Hence
$a_{1} =^{n} C_{1} = n$
$a_{2} = 2 +^{n} C_{2}$ .
$a_{3} =^{n} C_{3} + 2^{n} C_{1}$
Now
$2 a_{2} = a_{1} + a_{3}$
Or
$4 + 2^{n} C_{2} = 3^{n} C_{1} +^{n} C_{3}$
$4 + (n (n - 1)) = 3 n + \frac{n (n - 1) (n - 2)}{6}$
$4 - 3 n = n (n - 1) [\frac{n - 2}{6} - 1]$
$4 - 3 n = \frac{n (n - 1) (n - 8)}{6}$
$24 - 18 n = n (n^{2} - 9 n + 8)$
$n^{3} - 9 n^{2} + 8 n - 24 + 18 n = 0$
$n^{3} - 9 n^{2} + 26 n - 24 = 0$ .
Now sum of roots
$= \frac{- c o e f f i c i e n t o f x^{2}}{c o e f f i c i e n t o f x^{3}}$
$= 9$ .

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