Binomial expansion

**thereddevils** · February 25th 2009, 04:51 AM

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

**Danny** · February 25th 2009, 05:07 AM

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 $p^{2n}-2nq-1$ can be divided exactly by $q^2$ for all positive integers of n .

Using

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

we see your expression becomes

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

**thereddevils** · February 25th 2009, 05:20 AM

Originally Posted by danny arrigo

Using

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

we see your expression becomes

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

Thanks Danny , but i still don see how it works . Need a little more explaination .

Thanks again.

**thereddevils** · March 3rd 2009, 08:42 AM

Originally Posted by danny arrigo

Using

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

we see your expression becomes

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

oh i see now .. thanks danny

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