In the expansion of (a+1+frac{1}{a})^n,where | vedclass

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(C) The expression is $(a+1+\frac{1}{a})^n = \frac{(a^2+a+1)^n}{a^n}$.For a trinomial expansion $(x+y+z)^n$,the number of terms is given by $\frac{(n+

Vedclass · Accepted Answer

$1014$

Vedclass · Answer

(C) The expression is $(a+1+\frac{1}{a})^n = \frac{(a^2+a+1)^n}{a^n}$.For a trinomial expansion $(x+y+z)^n$,the number of terms is given by $\frac{(n+1)(n+2)}{2}$.However,in this specific case,the expression is $(a^2+a+1)^n$ divided by $a^n$. The expansion of $(a^2+a+1)^n$ has $2n+1$ terms because the powers of $a$ range from $a^0$ to $a^{2n}$.Given the number of terms is $2029$,we set $2n+1 = 2029$.$2n = 2028$.$n = 1014$.Hence,option $C$ is correct.

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

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