Find the constant coefficient of $(x+\frac{2}{x})^{100}$

Question

How do I find the constant coefficient for the following binomial?

$(x+ \frac{2}{x})^{100}$

I know a very similar question has been asked for $(x+ \frac{1}{x})^{100}$, but how does it differ for $(x+ \frac{2}{x})^{100}$ ?

Thank you so much in advance!

Answer 1

The only difference between finding the constant term coefficient of $(x+1/x)^{100}$ and that of $(x+2/x)^{100}$ is that you need to multiply the result for $(x+1/x)^{100}$ by $2^{50}$, because in $(x+2/x)^{100}$ the constant term consists of $50$ terms of $x$ and $50$ terms of $2/x$ being multiplied together.

With $(x+1/x)^{100}$ this extra factor can be ignored as $1^{50}=1$, but with $(x+2/x)^{100}$ this is not the case.

Answer 2

Let $T_{n+1}$ be the constant term. $$T_{n+1}=\binom{100}{n}x^{100-n}\big(\dfrac{2}{x}\big)^n$$ Now as $T_{n+1}$ is constant, $\dfrac{x^{100-n}}{x^n}$ is also constant, hence $n=50$

$\therefore$ Coefficient of the constant term is $$T_{51}=\binom{100}{50}2^{50}$$

Answer 3

$(a+ b)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}$.

Here, $a= x$, $b= 2/x$, and $n= 100$. The "constant term" is when $a^ib^{n-i}= x^i(1/x^{n-i})= x^{2i-n}= 1$ so when $i= n/2= 50$. The constant term is $\begin{pmatrix}100 \\ 50 \end{pmatrix}2^{50}$.