Find values: $ \sum_{k=1}^{15}k{15\choose k}\Big(\frac{1}{20}\Big)^k\Big(\frac{19}{20}\Big)^{15-k}$ and same one but with $k^2$

Question

$$ \sum_{k=1}^{15}k{15\choose k}\Big(\frac{1}{20}\Big)^k\Big(\frac{19}{20}\Big)^{15-k} $$ and $$ \sum_{k=1}^{15}k^2{15\choose k}\Big(\frac{1}{20}\Big)^k\Big(\frac{19}{20}\Big)^{15-k} $$

I do not know how to do it.

Answer 1

$$k^2{n\choose k}\Big(\frac{1}{20}\Big)^k\Big(\frac{19}{20}\Big)^{n-k}=\Big(\frac{19}{20}\Big)^nk^2\binom nk\left(\dfrac{\dfrac1{20}}{\dfrac{19}{20}}\right)^k=\Big(\frac{19}{20}\Big)^nk^2\binom nkx^k$$

where $x=\dfrac1{19}$

Now as $\displaystyle(1+x)^n=\sum_{k=0}^n x^k$

Differentiate both sides wrt $x,$ $$n(1+x)^{n-1}=\sum_{k=0}^nkx^{k-1}$$

Multiple both sides by $x$ and differentiate again

Answer 2

HINT:

$$(a+b)^n=\sum_{r=0}^n\binom nr a^rb^{n-r}$$

and for $\displaystyle k>0, k\cdot\binom nk=\cdots=n\binom{n-1}{k-1}$

For $\displaystyle k>1,$ $\displaystyle k^2\cdot\binom nk=\{k(k-1)+k\}\binom nk=n(n-1)\cdot\dfrac{(n-2)!}{(k-2)!\cdot\{n-2-(k-2)\}!}+n\cdot\binom{n-1}{k-1}=?$

Answer 3

The first one is a pretty easy, you can solve it like this:

$ \sum_{k=1}^{15}{k * \binom{k}{15} * (\frac{1}{20})^k*(\frac{19}{20})^{^{15-k}}}=\\ \sum_{k=1}^{15}{\frac{15!}{k!(15-k)!} *k *(\frac{1}{20})^k*(\frac{19}{20})^{15-k}}=\\ 15*\sum_{k=1}^{15}{\frac{14!}{(k-1)!(15-k)!}*(\frac{1}{20})^k*(\frac{19}{20})^{15-k}}=\\ 15*\sum_{k=0}^{14}{\frac{14!}{k!*(14-k)!}*(\frac{1}{20})^{k+1}*(\frac{19}{20})^{14-k}}=\\ 15*\frac{1}{20}*\sum_{k=0}^{14}{\binom{k}{14}(\frac{1}{20})^k*(\frac{19}{20})^{14-k}} =\\ \frac{15}{20}*(\frac{1}{20}+\frac{19}{20})^{14}=\frac{3}{4}\\$

The second one can be solved using the same methode.

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In the last line I used this equation:

$\sum_{k=0}^{n}{\binom{n}{k}*a^k*b^{n-k}} = (a + b)^n$