Can anyone help me understand this sum? sum from k=0 to 6 of k(6 choose k)(1/6)^k(5/6)^(6-k). My professor said that we should try to solve it without simply calculating each and every term.

Question

anonymous · Answer

$$\sum_{k=0}^{6}k \left(\begin{matrix}6 \ k\end{matrix}\right)(\frac{ 1 }{ 6 })^k(\frac{ 5 }{ 6 })^{(6-k)}$$

zarkon · Answer

$$\sum_{k=0}^{n}k{n\choose k}\left(p\right)^{k}\left(1-p\right)^{n-k}$$

$$=\sum_{k=1}^{n}k{n\choose k}\left(p\right)^{k}\left(1-p\right)^{n-k}$$

$$=\sum_{k=1}^{n}k\frac{n!}{k!(n-k)!}\left(p\right)^{k}\left(1-p\right)^{n-k}$$

$$=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}\left(p\right)^{k}\left(1-p\right)^{n-k}$$

$$=\sum_{k=1}^{n}\frac{n(n-1)!}{(k-1)!(n-k)!}\left(p\right)^{k}\left(1-p\right)^{n-k}$$

$$=\sum_{k=1}^{n}\frac{n(n-1)!}{(k-1)!(n-k)!}p\left(p\right)^{k-1}\left(1-p\right)^{n-k}$$

$$=np\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}\left(p\right)^{k-1}\left(1-p\right)^{n-k}$$
let $m=k-1$


$$=np\sum_{m=0}^{n-1}\frac{(n-1)!}{m!(n-k)!}\left(p\right)^{m}\left(1-p\right)^{n-k}$$

$$=np\sum_{m=0}^{n-1}\frac{(n-1)!}{m!(n-1-(k-1))!}\left(p\right)^{m}\left(1-p\right)^{n-1-(k-1)}$$

$$=np\sum_{m=0}^{n-1}\frac{(n-1)!}{m!(n-1-m)!}\left(p\right)^{m}\left(1-p\right)^{n-1-m}$$

$$=np\sum_{m=0}^{n-1}{n-1\choose m}\left(p\right)^{m}\left(1-p\right)^{n-1-m}$$


$$=np\cdot 1=np$$

for you $n=6$ and $p=1/6$ thus $np=1$