Binomial series with coeficients in arithmetic progression

[Appleton]

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}+...<br />

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

Homework Equations

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}\\<br /> \frac{\binom{n}{s-1} + \binom{n}{s+1}}{2}=\binom{n}{s} \\<br /> \frac{n!}{2(n-s+1)!(s-1)!} + \frac{n!}{2(n-s-1)!(s+1)!} = \frac{n!}{(n-s)!s!}\\<br /> \frac{(n-s)!s!}{(n-s+1)!(s-1)!} + \frac{(n-s)!s!}{(n-s-1)!(s+1)!} = 2\\<br /> \frac{s}{n-s+1} + \frac{n-s}{s+1}= 2\\<br /> \frac{n-s}{s+1} = 2- \frac{s}{n-s+1}\\<br /> n+2= (s+1)(2- \frac{s}{n-s+1})+s+2\\<br /> n+2= 2(s+1)- \frac{s(s+1)}{n-s+1}+s+2<br />
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]

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}+...<br />

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

Homework Equations

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}\\<br /> \frac{\binom{n}{s-1} + \binom{n}{s+1}}{2}=\binom{n}{s} \\<br /> \frac{n!}{2(n-s+1)!(s-1)!} + \frac{n!}{2(n-s-1)!(s+1)!} = \frac{n!}{(n-s)!s!}\\<br /> \frac{(n-s)!s!}{(n-s+1)!(s-1)!} + \frac{(n-s)!s!}{(n-s-1)!(s+1)!} = 2\\<br /> \frac{s}{n-s+1} + \frac{n-s}{s+1}= 2\\<br /> \frac{n-s}{s+1} = 2- \frac{s}{n-s+1}\\<br /> n+2= (s+1)(2- \frac{s}{n-s+1})+s+2\\<br /> n+2= 2(s+1)- \frac{s(s+1)}{n-s+1}+s+2<br />
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.

 

[Appleton]

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