Write the number of terms in the expansion of

To find: the number of terms in the expansion of $(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$

Formula Used:

Binomial expansion of $(x+y)^{n}$ is given by,

$(\mathrm{x}+\mathrm{y})^{\mathrm{n}}=\sum_{r=0}^{n}\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{x}^{\mathrm{n}-\mathrm{r}} \times \mathrm{y}^{\mathrm{r}}$

Thus,

$(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$

$=\left((\sqrt{2})^{5}+(\sqrt{2})^{4}\left(\begin{array}{l}5 \\ 1\end{array}\right)+\cdots+\left(\begin{array}{l}5 \\ 5\end{array}\right)\right)$

$+\left((\sqrt{2})^{5}-(\sqrt{2})^{4}\left(\begin{array}{l}5 \\ 1\end{array}\right)+\cdots-\left(\begin{array}{l}5 \\ 5\end{array}\right)\right)$

So, the no. of terms left would be 6

Thus, the number of terms in the expansion of $(\sqrt{2}+1)^{5}+(\sqrt{2}-1)^{5}$ is 6