Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is 
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"

Let $c_k$ be the coefficient of $\left(x+\dfrac{1}{x}\right)^{10}$ before $x^k$. Then
$$(x^2+1)\left(x+\dfrac{1}{x}\right)^{10} = (x^2+1)\sum\limits_{-\infty}^{+\infty} c_kx^k = \sum\limits_{-\infty}^{+\infty} c_kx^{k+2} + \sum\limits_{-\infty}^{+\infty} c_kx^{k} = \sum\limits_{-\infty}^{+\infty} (c_{k-2} + c_k)x^{k}$$
Since we are looking for the coefficient before $x^0$, the answer to your question is $c_{-2} + c_{0}$. 
According to binomial formula, we have:
$$\left(x+\dfrac{1}{x}\right)^{10} = \sum\limits_{k=0}^{10} \binom{10}{k}x^{k}\left(\dfrac{1}{x}\right)^{10-k} = \sum\limits_{k=0}^{10} \binom{10}{k}x^{2k - 10}$$
Thus,
$$c_{-2} = \{2k-10=-2 \Rightarrow k = 4\} = \binom{10}{4}$$
$$c_0 = \{2k-10=0 \Rightarrow k = 5\} = \binom{10}{5}$$
So, the answer is $$c_{-2} + c_0 = \binom{10}{5} + \binom{10}{4} = \binom{11}{5} = 462$$
