I don't have any idea to solve this problem. Find the value of
$$\sum_{k=n}^{2n} \binom{k}{n} 2^{-k}$$
For every integers $n\geq1$
Could you give me a hint to solve it? Once I get a hint, I will try to solve it again. Thanks
Asked
Active
Viewed 82 times
1
-
6You haven't tried anything, you haven't even calculated a few values (for $n=1,2,3,$ say) to get a notion what you're dealing with. – Feb 11 '18 at 09:01
-
1I second what Professor Vector said. Once you have an idea of what this sum yields and you try to prove it via induction, it might be helpful to see that $\sum_{k=n}^{2n}\begin{pmatrix}k\n\end{pmatrix}2^{-k}=\sum_{k=0}^{n}\begin{pmatrix}k+n\n\end{pmatrix}2^{-k-n}=2^{-n}\sum_{k=0}^{n}\begin{pmatrix}k+n\k\end{pmatrix}2^{-k}$ (simple index shift) – Frederik vom Ende Feb 11 '18 at 09:08
-
1Ok. The value is 1, and to prove that the value is 1, I should use induction. – Agung Izzul Haq Feb 11 '18 at 09:40
-
It's basically this after some simple substitutions Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ – Sil Feb 11 '18 at 10:04
-
You can also look here. – Feb 11 '18 at 10:44