I have problem with finding sum of series:
1)$\displaystyle\sum_{n=k}^{2k-1}\frac{n}{2^n}=?$
2) $\displaystyle\sum_{n=0}^{k-1}n(\frac{4}{3})^n=?$
I have some idea to 1) ot write it as $\displaystyle\frac{k\cdot2^{k-1}+(k+1)\cdot2^{k-2}+...+2k-1}{2^{2k-1}}=\frac{k(2+2^2+...+2^{k-1})+2^{k-2}+3\cdot2^{k-3}+..-1}{2^{2k-1}}$
but I don't know how to find sum of $2^{k-2}+3\cdot2^{k-3}+..-1$
Note that $$ 1+x+x^2+x^3+\ldots +x^k=\frac{1-x^{k+1}}{1-x}, \quad x \neq1, $$ gives by differentiation $$ 1\cdot x+2 \cdot x+3 \cdot x^2+\ldots +k \cdot x^{k-1}=\sum_{n=1}^{k-1}nx^n=-\frac{(k+1) x^k}{1-x}+\frac{1-x^{k+1}}{(1-x)^2}, \quad x \neq1. \tag1 $$
Hence
1) Write $\quad \displaystyle\sum_{n=k}^{2k-1}\frac{n}{2^n}=\sum_{n=1}^{2k-1}\frac{n}{2^n}-\sum_{n=1}^{k-1}\frac{n}{2^n}$ and use $(1)$ with $x=\dfrac12$.
2) For $\qquad\displaystyle\sum_{n=0}^{k-1}n\left(\frac{4}{3}\right)^n$ use $(1)$ with $x=\dfrac43$.
prove by induction that $\sum_{n=k}^{2k-1}\frac{n}{2^n}=-4\, \left( 1/2 \right) ^{2\,k}k-2\, \left( 1/2 \right) ^{2\,k}+2\, \left( 1/2 \right) ^{k}k+2\, \left( 1/2 \right) ^{k} $ for $\sum_{n=0}^{k-1}n\left(\frac{4}{3}\right)^n=3\,k \left( 4/3 \right) ^{k}-12\, \left( 4/3 \right) ^{k}+12$
