Binomial Expansion

**mathguy80** · May 3rd 2011, 05:55 AM

Hi Guys,

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

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

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

Thanks for your help.

**mr fantastic** · May 3rd 2011, 07:02 AM

Originally Posted by mathguy80

Hi Guys,

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

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

I tried putting the substitution for $a$ into the equation, and did a binomial expansion of $(b - 1)^{2n}$ , 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 $(b - 1)^{2n}$ . Then add 2nb and subtract 1. You can't help but notice something ....

**Moo** · May 3rd 2011, 07:06 AM

Hello,

Some steps have been volutarily skipped.

let's expand $(b-1)^{2n}$

$(b-1)^{2n}=\sum_{k=0}^{2n} C_{2n}^k b^k (-1)^{2n-k}=\sum_{k=0}^{2n} C_{2n}^k (-b)^k$

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

Hence $(b-1)^{2n} \bmod b^2 = \sum_{k=0}^1 C_{2n}^k (-b)^k \bmod b^2 = 1+2n(-b) \bmod b^2$

and it gives your result

**melese** · May 3rd 2011, 07:16 AM

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

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

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

**mathguy80** · May 3rd 2011, 09:14 AM

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!

**Moo** · May 3rd 2011, 10:40 AM

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

**mathguy80** · May 3rd 2011, 09:18 PM

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

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