# Thread: Binomial expansion

1. ## Binomial expansion

Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression $\displaystyle p^{2n}-2nq-1$ can be divided exactly by $\displaystyle q^2$ for all positive integers of n .

2. Originally Posted by thereddevils
Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression $\displaystyle p^{2n}-2nq-1$ can be divided exactly by $\displaystyle q^2$ for all positive integers of n .
Using

$\displaystyle (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$

we see your expression becomes

$\displaystyle (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$

3. Originally Posted by danny arrigo
Using

$\displaystyle (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$

we see your expression becomes

$\displaystyle (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$
Thanks Danny , but i still don see how it works . Need a little more explaination .

Thanks again.

4. Originally Posted by danny arrigo
Using

$\displaystyle (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$

we see your expression becomes

$\displaystyle (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots$

oh i see now .. thanks danny