Binomial expansion up to $x^{2}$

Question

The question is expand $(1+x+2ax^{2})^{b}$ in ascending power of $x$ up to the term in $x^{2}$.

I did as following:

$\bigg(1+\big(x+2ax^{2}\big)\bigg)^{b}=\left(^{b}_{0}\right)(1)^{b}+\left(^{b}_{1}\right)(1)^{b-1}(x+2ax^{2})+\left(^{b}_{2}\right)(1)^{b-2}(x+2ax^{2})^{2}+\dots$

$\hspace{37.5 mm}=(1)(1)+(b)(1)(x+2ax^{2})+\left(^{b}_{2}\right)(1)(x^{2}+4ax^{3}+4a^{2}x^{4})+\dots$

$\hspace{37.5 mm}=1+bx+2abx^{2}+\left(^{b}_{2}\right)x^{2}+\dots$

$\hspace{37.5 mm}=1+bx+\big(2ab+\left(^{b}_{2}\right)\big)x^{2}+\dots$

But the answer in my book has $2a$ not $2ab$.

Where did I make a mistake?

Answer 1

I think that you are right and the book is wrong.

For one thing, the coefficients are always going to have $b$ as part of them.

Discuss this with your instructor. The other students may also be worried.

Answer 2

As already said, the book is wrong.

Just for your curiosity, you could have done the work probably simpler using Taylor series $$y=(1+x+2ax^{2})^{b}\implies \log(y)=b\log(1+x+2ax^{2})$$ which already shows that $b$ will be present everywhere.

Now, using, around $t=0$ $$\log(1+t)=t-\frac{t^2}{2}+O\left(t^3\right)$$ and replacing $t=x2+2ax^2$ would lead to $$\log(1+x+2ax^{2})=x+\left(2 a-\frac{1}{2}\right) x^2+O\left(x^3\right)$$ $$\log(y)=bx+b\left(2 a-\frac{1}{2}\right) x^2+O\left(x^3\right)$$ and Taylor again $$y=e^{\log(y)}=1+b x+\frac{b}{2} \left(4 a +b-1\right)x^2 +O\left(x^3\right)$$ which is what you very correctly obtained.