In the expansion of $(1+x+x^2)^{20}$, find the number of terms in the binomial expansion.
Let $(1+x)$ be one term and $x^2$ as the second terms
$$\binom {20} {0} C x^{40}(1+x)^0+\binom {20} {1}Cx^{39}(1+x)^1+\dotsb+\binom {20} {20} x^0(1+x)^{20}$$
Number of terms $= 1+2+3+\dotsb +20 = 20\cdot \frac{21}{2}=210$?
But the answer is 41. How? Can anyone explain it to me why I am wrong? And why answer should be 41
This is a gmat exam question.
Best Answer
Indeed as the above comment by Xander Henderson explains, the degree of the polynomial is 40, hence it has at most 41 terms. In fact, it has exactly 41 terms as none of the coefficients of $x^i \ : \ i \in \{0, 1, 2, \cdots , 40\}$ are $0$.
In your method, you seemed to have missed that there are 210 terms before simplifying the expression, but it would seem that after collecting each term the final answer comes out to be 41.