Number of terms in the expansion of $\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$ 
Number of terms in the expansion of $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n$$ 

$\bf{My\; Try::}$ We can write $$\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^n=\frac{1}{x^{2n}}\left(1+x+x^2\right)^n$$
Now we have to calculate number of terms in $$(1+x+x^2)^n$$ and divided each term by $x^{2n}$
So $$(1+x+x^2)^n = \binom{n}{0}+\binom{n}{1}(x+x^2)+\binom{n}{2}(x+x^2)^2+......+\binom{n}{n}(x+x^2)^n$$
and divided each term by $x^{2n}$
So we get number of terms $\displaystyle  = 1+2+3+.....+n+1 = \frac{(n+1)(n+2)}{2}$
Is my solution is Right, If not then how can i calculate it
actually i don,t have solution of that question.
Thanks

When you expand $(1 + x + x^2)^n$, the highest degree is $2n$, and the lowest is 0. None of them is vanished, and it can be seen easily that there is one term of any degree between 0 and $2n$. So there should be $2n + 1$ terms.
Cheers,
