# Binomial series with coeficients in arithmetic progression

## Homework Statement

The binomial expansion of $(1+x)^n$, n is a positive integer, may be written in the form

$(1+x)^{n} = 1+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+...c_{r}x^{r}+...$

Show that , if $c_{s-1}$, $c_{s}$ and $c_{s+1}$ are in arithmetic progression then $(n-2s)^{2} =n+2$

## The Attempt at a Solution

"$c_{s-1}$, $c_{s}$ and $c_{s+1}$ are in arithmetic progression" infers that
$\frac{c_{s-1} +c_{s+1}}{2}=c_{s}\\ \frac{\binom{n}{s-1} + \binom{n}{s+1}}{2}=\binom{n}{s} \\ \frac{n!}{2(n-s+1)!(s-1)!} + \frac{n!}{2(n-s-1)!(s+1)!} = \frac{n!}{(n-s)!s!}\\ \frac{(n-s)!s!}{(n-s+1)!(s-1)!} + \frac{(n-s)!s!}{(n-s-1)!(s+1)!} = 2\\ \frac{s}{n-s+1} + \frac{n-s}{s+1}= 2\\ \frac{n-s}{s+1} = 2- \frac{s}{n-s+1}\\ n+2= (s+1)(2- \frac{s}{n-s+1})+s+2\\ n+2= 2(s+1)- \frac{s(s+1)}{n-s+1}+s+2$
At this point it's getting messy, so I set s to 1 and n to 2 and get n+2 = 6 as opposed to the 0 I get when I do the same susubstitution for $(n-2s)^{2}$
Can someone please point out what I have done wrong.

haruspex
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## Homework Statement

The binomial expansion of $(1+x)^n$, n is a positive integer, may be written in the form

$(1+x)^{n} = 1+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+...c_{r}x^{r}+...$

Show that , if $c_{s-1}$, $c_{s}$ and $c_{s+1}$ are in arithmetic progression then $(n-2s)^{2} =n+2$

## The Attempt at a Solution

"$c_{s-1}$, $c_{s}$ and $c_{s+1}$ are in arithmetic progression" infers that
$\frac{c_{s-1} +c_{s+1}}{2}=c_{s}\\ \frac{\binom{n}{s-1} + \binom{n}{s+1}}{2}=\binom{n}{s} \\ \frac{n!}{2(n-s+1)!(s-1)!} + \frac{n!}{2(n-s-1)!(s+1)!} = \frac{n!}{(n-s)!s!}\\ \frac{(n-s)!s!}{(n-s+1)!(s-1)!} + \frac{(n-s)!s!}{(n-s-1)!(s+1)!} = 2\\ \frac{s}{n-s+1} + \frac{n-s}{s+1}= 2\\ \frac{n-s}{s+1} = 2- \frac{s}{n-s+1}\\ n+2= (s+1)(2- \frac{s}{n-s+1})+s+2\\ n+2= 2(s+1)- \frac{s(s+1)}{n-s+1}+s+2$
At this point it's getting messy, so I set s to 1 and n to 2 and get n+2 = 6 as opposed to the 0 I get when I do the same susubstitution for $(n-2s)^{2}$
Can someone please point out what I have done wrong.
It's no use checking the equation by using a combination of values that is not actually a solution. n=7, s=2 is a solution, and your last equation survives that test.
Go back to your 5th equation (the one with just =2 on the right hand side) and just multiply out the denominators.

OK, yes, pretty fundamental misunderstanding then. Thanks for pointing that out.