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  • Let n be a positive integer and <img alt="\left(1+x+x^2\right)^n=a_0+a_1 x+\ldots+a_{2 n} x^{2 n}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%281&plus;x&plus;x%5E2%5C

Let n be a positive integer and \left(1+x+x^2\right)^n=a_0+a_1 x+\ldots+a_{2 n} x^{2 n}, then a_0{ }^2-a_1^2+\ldots .+a_{2 n}^2 equals to

Option: 1

a_n


Option: 2

a_0+a_n


Option: 3

n a_n


Option: 4

None of these


Answers (1)

best_answer

\text{Given}\ \left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}$\\\ Replacing $x$ by $-\frac{1}{x}$ in (1), we get $$ \left(1-\frac{1}{x}+\frac{1}{x^2}\right)^n=a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+\ldots . . .+\frac{a_{2 n}}{x^{2 n}} $$ Multiplying (1) and (2), we get $$ \begin{gathered} \frac{\left(1+x+x^2\right)^n\left(1-x+x^2\right)^n}{x^{2 n}}=\left(a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}\right) \\ \times\left(a_0-\frac{a_1}{x}+\frac{a_2}{x^2}+\ldots \ldots+\frac{a_{2 n}}{x^{2 n}}\right) \end{gathered}

Term independent of x on the R.H.S.
=a_0^2-a_1^2+a_2^2-\ldots .+a_{2 n}^2 \text {. }

It is equal to the term independent of x in L.H.S.

\begin{aligned} & \quad=\text { the coefficient of } x^{2 n} \text { in L.H.S } \\ & =\text { the coefficient of } x^{2 n} \text { in }\left(1+x^2+x^4\right)^n \\ & =\text { the coefficient of } t^n \text { in }\left(1+t+t^2\right)^n \quad\left[\text { Putting } x^2=t\right] \\ & \therefore a_0^2-a_1{ }^2+a_2^2-a_3^2+\ldots . .+a_{2 n}{ }^2=a_n . \quad[\text { By }(1)] \end{aligned}

Posted by

Irshad Anwar

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