#2 08. 12. 2018 18:57 — Editoval laszky (08. 12. 2018 20:49)
Re: Binomial sum
, we get![$S_{n+1} = \sum^{n+1}_{r=0}\binom{n+1+r}{n+1} \left(\frac{1}{2}\right)^{r} = 1+\sum^{n+1}_{r=1}\left[\binom{n+r}{n}+\binom{n+r}{n+1}\right] \left(\frac{1}{2}\right)^{r} = $](/mathtex/91/91efb1a2627040c348e1658d5200e210.gif "kopírovat do textarea")
![$ = 1+\left[ \sum^{n}_{r=0}\binom{n+r}{r} \left(\frac{1}{2}\right)^{r} + \binom{2n+1}{n} \left(\frac{1}{2}\right)^{n+1}-1\right] + \sum^{n}_{r=0}\binom{n+1+r}{n+1} \left(\frac{1}{2}\right)^{r+1} = $](/mathtex/49/4949080e6378ea110012062dddcb77f9.gif "kopírovat do textarea")
![$S_{n+1} = S_n+\frac{1}{2}S_{n+1} + \left[2\binom{2n+1}{n}-\binom{2n+2}{n+1}\right]\left(\frac{1}{2}\right)^{n+2}$](/mathtex/21/218084ae0b63dbebd9338fa3addafe4b.gif "kopírovat do textarea")
.
, which means that
and since 
, we get 
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#3 09. 12. 2018 13:12
- stuart clark
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Re: Binomial sum
Thanks ↑ laszky: for nice explanation.
Can we solve it using combinational argument or using probability argument (like tossing a coin or something.)
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