# What is the coefficient of $x^{50}$ in $\left(x + \frac1x\right )^{100}$?

I know I should use the Binomial Theorem, but I'm just having some trouble figuring this out. thanks!

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Hint: $$\left(x+\frac1x\right)^{100} = \sum_{k=0}^{100} {100\choose k} x^{k}\frac{1}{x^{100-k}} = \sum_{k=0}^{100} {100\choose k} x^{k}x^{k-100} = \sum_{k=0}^{100} {100\choose k} x^{2k-100}.$$

Now, just find the values of $k$ for which $2k-100 = 50$.

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The coefficient of $x^{50}$ in $\displaystyle\left(x+\frac1x\right)^{100}=\frac{(1+x^2)^{100}}{x^{100}}$

$=$ the coefficient of $x^{100+50}$ in $\displaystyle(1+x^2)^{100}$

Now, the coefficient of $x^{2r}$ in $\displaystyle(1+x^2)^{100}$ is $\displaystyle\binom{100}r$

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$$(x+\frac{1}{x})^{100} =\sum_{} {n\choose r}x^r(x^{-1})^{100-r}$$

Solving $r - (100 - r) = 50 \implies r=75$

Thus coefficient is ${100 \choose 75}$

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You can also do this without the binomial theorem (actually, we are using the reason behind the binomial theorem; so in this method you may gain some intuition about the binomial theorem itself!).

In order to expand $$\left(x+\frac{1}{x}\right)^{100}=\left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\cdots \left(x+\frac{1}{x}\right)$$ you have to write down all the possible choices of $x$ or $1/x$ in each term of the product. Each of those choices that pick $x$ in $a$ terms, and $1/x$ in $b$ terms will give you a term $x^a\cdot (1/x)^b=x^{a-b}$. Since there are $100$ terms, we have $a+b=100$, so we may write our term as $x^{2a-100}$ instead. If $x^{2a-100}=x^{50}$, then $a=75$. It remains to find out how many times we can have picked out $x$ in $75$ terms of the product. This boils down to selecting $75$ terms out of $100$ terms to be the ones where we pick $x$. That is, in how many ways we can pick $75$ items out of $100$? This is, by definition, the combinatorial number $\binom{100}{75}$.

Hence, we conclude that there will be $\binom{100}{75}$ terms in the product with a $x^{50}$, and therefore the coefficient of $x^{50}$ in the expansion is precisely $\binom{100}{75}$.

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