Let 1- x 2 2 - 1- x n - k-0 n-4 a k - x k- If a 1 - a 2 and a 3 are in A-P- find n-

(1+x^2)^2(1+x)^n =(1+2x^2+x^4)(1+nx+ ^nC_2x^2...) Hence #160; a_1= ^nC_1=n a_2=2+ ^nC_2 . a_3= ^nC_3+2 ^nC_1 Now #160; 2a_2=a_1+a ...

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Binomial Coefficients

Let 1+ x 2 ...

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 $n$ .

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(1 + x^{2})^{2} (1 + x)^{n} = n + 4 \sum k = 0 a_{k} x_{k}

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{(1 + x^{2})}^{2} {(1 + x)}^{n} = \sum_{k = 0}^{n + 4} a_{k} x^{k}

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(1 + x^{2})^{2} (1 + x)^{n} = n + 4 \sum k = 0 a_{k} x^{k} .

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