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}}
$"
 A: By writing the expression as $$\frac{(x^2+1)^{11}}{x^{10}}$$
It is clear that we need the term involving $x^{10}=(x^2)^5$ in the expansion of $(x^2+1)^{11}$ which is $$\binom{11}{5}$$
A: $f(x)=(x^2+1)(x+\dfrac{1}{x})^{10}$
We can rewrite $f(x)$ like below:
$f(x) = x^2(x+\dfrac{1}{x})^{10} + (x+\dfrac{1}{x})^{10}$
In the first term, the power of $x$, must be $-2$ in the parenthesis, so the when it's multiplyed by $x^2$  the power of $x$ becomes $0$.    And we know that:
$(x+\dfrac{1}{x})^{10}=\sum_{k=0}^{10}\binom{10}{k}x^k(\dfrac{1}{x})^{10-k}$
So $k+k-10=-2=>k=4$. So the coefficient of that term is $\binom{10}{4}$.
On the other hand, the coefficient of the constant term of the second term of $f(x)$ which is just $(x+\dfrac{1}{x})^{10}$, is when $k=5$. So the coefficient would be $\binom{10}{5}$
So the answer is:
$\binom{10}{4}+\binom{10}{5}$
A: We want to find the coefficients of the binomial expansion of $(x+\frac{1}{x})^{10}$ of the constant term and of the term with $\frac{1}{x^2}$.
$$\left(x+\frac{1}{x}\right)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{10-k}\frac{1}{x^k}$$
To find the constant term, want $10-k=k$, ie: $k=5$. This term is $\binom{10}{5}x^5\cdot x^{-5}=\binom{10}{5}$. To find the $\frac{1}{x^2}$ term, want $10-2k=2$, ie: $k=4$. Doing the same as above, get that this term is $\binom{10}4 x^{-2}$.
So when you multiply it out, you get that the constant term is $\binom{10}5+\binom{10}4$.
A: 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 $$c_{-2} + c_0 = \binom{10}{5} + \binom{10}{4} = \binom{11}{5} = 462$$
