Give a formula for the coefficient of $x^{k}$ in the expansion of $(x+{1 \over x})^n$

Question

I followed the binomial theorem and got this:

The Binomial Theorem is: $(a+b)^{n}= \sum_{k=0}^{n} {n \choose k}{a}^{k}{b}^{n-k}$

Then let $a=x, b={1\over x}, n = n, k = k.$

I then get $\sum_{j=0}^{n}{n \choose j}{x}^{j}{1 \over x}^{n-j}$

I'm not sure if I'm doing it right. Can someone help me out with this?

Change the index of the sum to some other variable say $j$, instead of $k$. — Anurag A, Feb 03 '16 at 09:30
Now the power of $x$ in a general term is given by $2j-n$. You want a $j$ such that $2j-n=k$. — Anurag A, Feb 03 '16 at 09:33

score 2 · Answer 1 · answered Feb 03 '16 at 09:39

\begin{align*} \left( x+\frac{1}{x} \right)^{n} &= x^{-n}\left( x^{2}+1 \right)^{n} \\ &= x^{-n}\sum_{j=0}^{n} \binom{n}{j} x^{2j} \\ &= \sum_{j=0}^{n} \binom{n}{j} x^{2j-n} \end{align*}

\begin{align*} k &= 2j-n \\ c_{k} &= \frac{1+(-1)^{n+k}}{2} \binom{n}{\frac{n+k}{2}} \end{align*}

score 1 · Answer 2 · answered Feb 03 '16 at 09:34

You're on the right track. You see $$(x+\tfrac 1x)^n=\sum_{k=0}^{n}{n \choose k}{x}^{k}{1 \over x}^{n-k}$$ Since $\frac 1x=x^{-1}$ we have $$(x+\tfrac 1x)^n=\sum_{k=0}^{n}{n \choose k}{x}^{k}x^{k-n}=\sum_{k=0}^{n}{n \choose k}{x}^{2k-n}$$ So if we wish to find the coefficient $a_i$ for $x^i$, then $a_i=0$ if $i\not\equiv n \mod 2$ and $a_i={n\choose \frac 12 (i+n)}$ otherwise.

Give a formula for the coefficient of $x^{k}$ in the expansion of $(x+{1 \over x})^n$

2 Answers2