# Sum

• August 6th 2009, 08:41 AM
Sum
How to prove that $\sum_{m=0}^n\binom{2n}{2m}9^{n-m}=\frac12\big(4^{2n}+2^{2n}\big)$ ?
• August 6th 2009, 09:32 AM
red_dog
Let $S_1=\sum_{m=0}^nC_{2n}^{2m}9^{n-m}=\sum_{m=0}^nC_{2n}^{2n}3^{2n-2m}$

$S_2=\sum_{m=1}^nC_{2n}^{2m-1}3^{2n-2m+1}$

Then

$S_1+S_2=(3+1)^{2n}=4^{2n}$

$S_1-S_2=(3-1)^{2n}=2^{2n}$

$2S_1=4^{2n}+2^{2n}\Rightarrow S_1=\frac{1}{2}(4^{2n}+2^{2n}$
• August 6th 2009, 12:24 PM
Thanks but, I don't see how did you get $S_1+S_2=(3+1)^{2n}$ ? I can't follow that equality, I was trying to figure out by myself but I couldn't.
• August 6th 2009, 01:08 PM
Amer
Quote:

Thanks but, I don't see how did you get $S_1+S_2=(3+1)^{2n}$ ? I can't follow that equality, I was trying to figure out by myself but I couldn't.

$S_1+S_2 =\sum_{m=0}^{n} \left(\begin{array}{c}2n\\2m\end{array}\right) 3^{2n-2m} + \sum_{m=1}^n \left(\begin{array}{c}2n\\2m-1\end{array}\right) 3^{2n-2m+1}$

lats expand S1

$\sum_{m=0}^{n} \left(\begin{array}{c}2n\\2m\end{array}\right) 3^{2n-2m} = 3^{2n} + \left(\begin{array}{c}2n\\2\end{array}\right)3^{2n-2} + \left(\begin{array}{c}2n\\4\end{array}\right) 3^{2n-4} + ... +\left(\begin{array}{c}2n\\2n-2\end{array}\right)3^2$

S2 in the same way

$\sum_{m=1}^n \left(\begin{array}{c}2n\\2m-1\end{array}\right) 3^{2n-2m+1} = \left(\begin{array}{c}2n\\1\end{array}\right)3^{2n-1} + \left(\begin{array}{c}2n\\3\end{array}\right)3^{2n-3} +... + \left(\begin{array}{c}2n\\2n-1\end{array}\right)3$

$S_1 + S_2 = 3^{2n} +\left(\begin{array}{c}2n\\1\end{array}\right)3^{2 n-1} + \left(\begin{array}{c}2n\\2\end{array}\right)3^{2n-2} +$ $\left(\begin{array}{c}2n\\3\end{array}\right)3^{2n-3} +\left(\begin{array}{c}2n\\4\end{array}\right) 3^{2n-4}$ $+...+ \left(\begin{array}{c}2n\\2n-2\end{array}\right)3^2+ \left(\begin{array}{c}2n\\2n-1\end{array}\right)3$

this is the the expand of

$(1+3)^{2n}$