Determine the coefficient of $x^{18}$ in $\left(x+\frac{1}{x}\right)^{50}$

Question

Determine the coefficient of $x^{18}$ in $\left(x+\frac{1}{x}\right)^{50}$.

I know he Binomial Theorem will be useful here, but I am struggling to use it with any certainty.

Answer 1

Hint The general term of the binomial expansion is $${50 \choose k} x^k \left(\frac{1}{x}\right)^{50 - k} = {50 \choose k} x^{2k - 50} .$$

Answer 2

${50 \choose k} x^k x^{k-50}$.

You need $k+(k-50)=18$, therefore $2k-50=18$, and $k=34$.

Calculating ${50 \choose 34}=4923689695575$ - your answer

Answer 3

Since for $y=x^2$,

$$\left(x+\frac{1}{x}\right)^{50}=\frac{\left(x^2+1\right)^{50}}{x^{50}}=\frac{\left(y+1\right)^{50}}{y^{25}}$$

you are looking for the coefficient of $y^{9+25}=y^{34}$ in $(y+1)^{50}$.