In the expansion of $\left(a+1+\frac{1}{a}\right)^n$, where $n \in \mathbf{N}$ there are 2029 terms. Then $n=$

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Aicharya · Accepted Answer

$\left(a+1+\frac{1}{a}\right)^n=\frac{1}{a^n}\left(a^2+a+1\right)^n$ $\therefore$ Number of terms $=2 n+1$ $\begin{aligned} 2029 & =2 n+1 \ 2 n & =2028 \ n & =1014 \end{aligned}$ Hence, option (c) is correct.

In the expansion of \(\left(a+1+\frac{1}{a}\right)^n\), where \(n \in \mathbf{N}\) there are 2029 terms.…

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