Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$

Question

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}} $"

Answer 1

$f(x)=(x^2+1)(x+\dfrac{1}{x})^{10}$

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:

$\binom{10}{4}+\binom{10}{5}$

Answer 2

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 $\binom{10}5+\binom{10}4$.

Answer 3

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 $$c_{-2} + c_0 = \binom{10}{5} + \binom{10}{4} = \binom{11}{5} = 462$$

Answer 4

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 $$\binom{11}{5}$$