Thread: Binomial Expansion

Hi Guys,

Need a little help with this one, Not sure how to proceed.

Two positive integers, $\displaystyle a,b$ are connected by the relation $\displaystyle a = b - 1$. By using a binomial expansion show that for all positive integral values of n the expression $\displaystyle a^{2n} + 2nb - 1$ is exactly divisible by $\displaystyle b^2$. By choosing suitable values of a and n, show that $\displaystyle 2^{40} + 119$ is divisible by 9, and hence that $\displaystyle 2^{39} + 1$ is divisible by 9.

I tried putting the substitution for $\displaystyle a$ into the equation, and did a binomial expansion of $\displaystyle (b - 1)^{2n}$, but not sure where that is supposed to lead. It seems like another longer equation.

Thanks for your help.

Originally Posted by mathguy80

Hi Guys,

Need a little help with this one, Not sure how to proceed.

Two positive integers, $\displaystyle a,b$ are connected by the relation $\displaystyle a = b - 1$. By using a binomial expansion show that for all positive integral values of n the expression $\displaystyle a^{2n} + 2nb - 1$ is exactly divisible by $\displaystyle b^2$. By choosing suitable values of a and n, show that $\displaystyle 2^{40} + 119$ is divisible by 9, and hence that $\displaystyle 2^{39} + 1$ is divisible by 9.

I tried putting the substitution for $\displaystyle a$ into the equation, and did a binomial expansion of $\displaystyle (b - 1)^{2n}$, but not sure where that is supposed to lead. It seems like another longer equation.

Thanks for your help.

I suggest you write out the first few terms and the last few terms of $\displaystyle (b - 1)^{2n}$. Then add 2nb and subtract 1. You can't help but notice something ....

Hello,

Some steps have been volutarily skipped.

let's expand



Considering it mod b², we can remove the powers of b superior to 2, because if for example , b^2 can be factored out and since b^2=0 mod b^2, it cancels all the terms with k>2.

Hence

and it gives your result

The Binomial Theorem gives $\displaystyle (b-1)^{2n}=1-\binom{2n}{1}b+\binom{2n}{2}b^2+\cdots$. Note that $\displaystyle \binom{2n}{1}=2n$ and that all terms after the second are divisible by $\displaystyle b^2$. Thus, $\displaystyle (b-1)^{2n}=1-2nb+Ab^2$, where $\displaystyle A$ is some integer.

Therefore, $\displaystyle a^2+2nb-1=Ab^2$.

This is essentially the same idea as in Moo's post.

Thanks guys. @Moo, sorry to sound stupid. I got the summation part, but the rest of the notation went way over my head.

@melese. Got it! The 1 and 2nb terms cancel out, rest all terms have powers of b >= 2. Then rest a=2,b=3 and n=20, forms similar expression hence divisible, which checks out. Thanks!

What part exactly ? C_{2n^k} corresponds to
And the sum goes to 1 because after 2 it has no interest (=0)

Yup, I was confused by the C_ notation. So that corresponds to n choose k. That clears things up, thanks!