Silly mistake on evaluating the sixth term of $\left (\frac{a}{b}+\frac{b}{a^2}\right)^{17}$?

Question

I am trying to evaluate the sixth term of $\displaystyle \left (\frac{a}{b}+\frac{b}{a^2}\right)^{17}$ with the binomial theorem. I've done the following:

The sixth term might be the term for $k=5$ in $\displaystyle \sum_{k=0}^{n}{n \choose k}x^{n-k}y^k$, then:

$${ 17 \choose 5}* \left(\frac{a}{b}\right)^{12}* \left(\frac{b}{a^2}\right)^{5}=\frac{6188 a^{12}b^5}{b^{12}a^{10}}=\frac{6188a^2}{b^7}$$

But I may be doing some silly mistake because the answer is: $\displaystyle\frac{6188 b^7}{a^{19}}$.

Answer 1

The answer book may think the binomial expansion is $(x+y)^n=\displaystyle \sum_{k=0}^{n}{n \choose k}x^k y^{n-k}$ so you have given the sixth term starting from one end and it has given the sixth term starting at the other.

It is just a matter of convention (as is whether you start counting at $0$ or $1$) and neither is particularly wrong, though if forced to choose I would probably go with the answer book's choice.

Answer 2

It should be $\binom{17}{5}\cdot \left(\dfrac{a}{b}\right)^5\cdot \left(\dfrac{b}{a^2}\right)^{12}$