Question

In the binomial expansion of (a+b)^n the coefficients of the terms equidistant from the beginning and the ending are always..? I know the answer is EQUAL. But how?

Answer 1

See below:

Explanation:

Let's talk for a second about the formula for the binomial expansion. That formula is:

`(a+b)^n=(C_(n,0))a^nb^0+ (C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n`

The coefficients you are referring to are from the Combination term and there are a couple of ways to demonstrate that symmetry you are referring to.

First off, it'd be good to know that the general formula for a Combination is:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

Let's see what the series of Combinations are for `n=4, k=0,1,2,3,4`:

`C_(4,0)=(4!)/(0!4!)=(4!)/(4!)=1`

`C_(4,1)=(4!)/(1!3!)=(4!)/(3!)=4`

`C_(4,2)=(4!)/(2!2!)=(4!)/(4)=6`

`C_(4,3)=(4!)/(3!1!)=(4!)/(3!)=4`

`C_(4,4)=(4!)/(4!0!)=(4!)/(4!)=1`

It's all in the divisor of the Combination formula. I'll leave the explanation here but there is more to it to explore. This link might help:

https://en.wikipedia.org/wiki/Combination

A table form of the symmetry is provided by something called Pascal's Triangle, the first few rows which looks like this:

Look 5 rows down. Do you see the terms we calculated?

Each row can be calculated by maintaining the triangular shape and by adding the two numbers above it (any side that only has one number, and I'm now referring to the side edges, are presumed to have a 0 where there is no number).

There are many cool relations hidden inside this triangle and mathematicians are constantly finding more. I'm attaching a video from one of my favourite youtube channels, Numberphile, which has a discussion on this: