Binomial theorem

• January 22nd 2009, 02:48 AM
alexmahone
Binomial theorem
If $(1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n$, show that
$c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}$
• January 22nd 2009, 03:52 AM
PaulRS
That is equivalent to showing that: $
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 } = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}
$
(1)

To prove it: $
\left( {1 + x} \right)^n = \sum\limits_{k = 0}^n {c_k \cdot x^k } \underbrace \Rightarrow _{{\rm{differentiate}}}n \cdot \left( {1 + x} \right)^{n - 1} = \sum\limits_{k = 0}^n {c_k \cdot k \cdot x^{k - 1} }
$

$
\Rightarrow n \cdot x \cdot \left( {1 + x} \right)^{n - 1} = \sum\limits_{k = 0}^n {c_k \cdot k \cdot x^k }
$
$
\Rightarrow n \cdot x \cdot \left( {1 + x} \right)^{n - 1} + \left( {1 + x} \right)^n = \sum\limits_{k = 0}^n {c_k \cdot \left( {k + 1} \right) \cdot x^k }
$

Now consider the coefficient of $x^n$ in the product: $
\left( {\sum\limits_{k = 0}^n {c_k \cdot \left( {k + 1} \right) \cdot x^k } } \right) \cdot \left( {\sum\limits_{k = 0}^n {c_k \cdot x^k } } \right)
$

It is: $
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k \cdot c_{n - k} } = \sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 }
$
this last equality holds since: $c_{n-k}=c_k$

NOw, on the other hand: $
\left( {\sum\limits_{k = 0}^n {c_k \cdot \left( {k + 1} \right) \cdot x^k } } \right) \cdot \left( {\sum\limits_{k = 0}^n {c_k \cdot x^k } } \right) = \left( {n \cdot x \cdot \left( {1 + x} \right)^{n - 1} + \left( {1 + x} \right)^n } \right) \cdot \left( {1 + x} \right)^n
$
$
= n \cdot x \cdot \left( {1 + x} \right)^{2n - 1} + \left( {1 + x} \right)^{2n}
$

So from here we also want the coefficient of $x^n$, which is exactly the RHS of (1) -by using the binomial theorem-, now, from the uniqueness of the coefficients of the polynomials it follows that $
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 } = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}
$
• January 22nd 2009, 05:50 AM
pankaj
Quote:

Originally Posted by alexmahone
If $(1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n$, show that
$c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}$

Let 1) $S=c_0^2 + 2c_1^2 + ... + nc_{n-1}^2+(n + 1)c_n^2$

$S=(n+1)c_n^2+nc_{n-1}^2+...+2c_1^2+c_0^2$

Since. $c_r=c_{n-r},$we get

2) $S=(n+1)c_0^2+nc_1^2+...+2c_{n-1}^2+c_n^2$

Adding 1 and 2 we get

$2S=(n+2)(c_0^2+c_1^2+...+c_{n-1}^2+c_n^2)$

$
2S=(n+2)C(2n,n)=(n+2) \frac{2n!}{n!n!}
$
• January 22nd 2009, 06:01 AM
alexmahone
Quote:

Originally Posted by pankaj
$2S=(n+2)(c_0^2+c_1^2+...+c_{n-1}^2+c_n^2)$

$
2S=(n+2)C(2n,n)
$

I don't get this step; could you please explain?
• January 22nd 2009, 07:06 AM
pankaj
$c_0^2+c_1^2+c_2^2+.......+c_n^2$

$=c_0c_n+c_1c_{n-1}+c_2c_{n-2}+.......+c_nc_0$

This is number of ways to select n objects from 2n objects which is alternatively $C(2n,n).$