Coefficient of $x^2$ in $(x+\frac 2x)^6$

Question

I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?

Answer 1

You have mistake.

$\binom{6}{4} . x^2 . \left(\frac 2x\right)^4$

= $15 . x^2 . \frac{16}{x^4}$

= $240 . \frac{1}{x^2}$

Here 240 is coefficient of $x^{-2}$ not $x^2$.

Correct term -

$\binom{6}{2} . x^4 . \left(\frac 2x\right)^2$

= $15 . x^4 . \frac{4}{x^2}$

= $60 . x^2$

So answer is 60.

Answer 2

It's much easier [for me] to see where the $x^2$ terms are after rewriting: $$(x + \dfrac{2}{x})^6 = (\dfrac{x^2 + 2}{x})^6 = \dfrac{1}{x^6}(x^2 +2)^6$$

Then we only have $x^2$ terms where the expansion of the binomial gives $x^8$ (because we're dividing everything by $x^6$). That term is $15(x^2)^4 \cdot 2^2$.

Answer 3

$\binom{6}{4}x^2\times(\frac{2}{x})^4$ would give the coefficient for $\frac{1}{x^2}$. What you want instead is $$\binom{6}{2}x^4\times(\frac{2}{x})^2$$

Answer 4

We have $$\displaystyle\left(x+\dfrac{2}{x}\right)^6=\sum_{k=0}^{6}\binom{6}{k}x^{6-k}\left(\frac{2}{x}\right)^k=\sum_{k=0}^{6}2^k\binom{6}{k}x^{6-2k}.$$ Then, $6-2k=2\Leftrightarrow k=2,$ so $$\text{coef }x^2=2^2\binom{6}{2}=60.$$

Answer 5

In the expansion all the terms will be of the form $$\binom{6}{k}x^k\left(\frac{2}{x}\right)^{6-k}.$$ and we need $$x^k ~x^{k-6}=x^2$$ i.e$$k=4.$$ Hence the coefficient is$$4\binom{6}{4}=60$$