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I have got this series of binomial coefficients -

$${2n\choose 0}+3\times{2n\choose 2}+3^{2}\times{2n\choose 4}+\ldots +3^{n}\times{2n\choose 2n}$$

I have to prove this to be divisble by $2^{n}$. I tried applying binomial theorem but it didn't work out. I only proved that this sum will be even. Can anyone help?

AvZ
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As $$(a+b)^{2n}+(a-b)^{2n}=2\sum_{r=0}^n\binom{2n}{2r}a^{2n-2r}b^{2r},$$

$$2\sum_{r=0}^n\binom{2n}{2r}(\sqrt3)^{2r}=(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}=(4+2\sqrt3)^n+(4-2\sqrt3)^n$$

If $T_n=\dfrac{(4+2\sqrt3)^n+(4-2\sqrt3)^n}{2^n}=\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n$

So, $T_{n+2}-[(2+\sqrt3)+(2-\sqrt3)]T_{n+1}+(2+\sqrt3)(2-\sqrt3)T_n=0\iff T_{n+2}=4T_{n+1}-T_n$

We need to show that $T_n/2$ is an integer which can be easily done using strong induction

lab bhattacharjee
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