# Thread: last terms of expansion

1. ## last terms of expansion

$\displaystyle (q+p)^n - (q-p)^n = 2 ({n}C1) q^{n-1}p + 2 ({n}C3)q^{n-3}p^3 +.....$

1. what is the last term of the expansion if n is odd

2. what is the last term of the expansion if n is even

how do i find last terms?

$\displaystyle (q+p)^n - (q-p)^n = 2 ({n}C1) q^{n-1}p + 2 ({n}C3)q^{n-3}p^3 +.....$

1. what is the last term of the expansion if n is odd

2. what is the last term of the expansion if n is even

how do i find last terms?
You'll have realised, of course, that the 1st, 3rd, ... terms (the odd terms) in the expansions of $\displaystyle (q+p)^n$ and $\displaystyle (q-p)^n$ have been eliminated by subtraction, leaving us with the even terms, which are then added together to give the factor of $\displaystyle 2$ in each term in the final expression.

Then notice that in the expansion of $\displaystyle (q+p)^n$ there are $\displaystyle n+1$ terms altogether, the last two terms being
$\displaystyle ...+\underbrace{nqp^{n-1}}_{n^{th}\text{ term}}\quad + \underbrace{p^n}_{(n+1)^{th}\text{ term}}$
So if $\displaystyle n$ is odd, $\displaystyle n+1$ is even, and the final term will be $\displaystyle 2p^n$.

And if $\displaystyle n$ is even, $\displaystyle n+1$ is odd, and the final term will be $\displaystyle 2nqp^{n-1}$.

Hello purebladeknightYou'll have realised, of course, that the 1st, 3rd, ... terms (the odd terms) in the expansions of $\displaystyle (q+p)^n$ and $\displaystyle (q-p)^n$ have been eliminated by subtraction, leaving us with the even terms, which are then added together to give the factor of $\displaystyle 2$ in each term in the final expression.
Then notice that in the expansion of $\displaystyle (q+p)^n$ there are $\displaystyle n+1$ terms altogether, the last two terms being
$\displaystyle ...+\underbrace{nqp^{n-1}}_{n^{th}\text{ term}}\quad + \underbrace{p^n}_{(n+1)^{th}\text{ term}}$
So if $\displaystyle n$ is odd, $\displaystyle n+1$ is even, and the final term will be $\displaystyle 2p^n$.
And if $\displaystyle n$ is even, $\displaystyle n+1$ is odd, and the final term will be $\displaystyle 2nqp^{n-1}$.