Question

Compute the summation ${\sum }_{k=0}^{27}k{(}_k^{27}\left)\right(\frac{1}{2}{)}^k(\frac{2}{3}{)}^{27-k}$

Answer 1

${\sum }_{k=0}^{27}k{}^{27}{C}_k{\left(\frac{1}{2}\right)}^k{\left(\frac{2}{3}\right)}^{27-k}$ $={\sum }_{k=1}^{27}27{}^{26}{C}_{k-1}{\left(\frac{1}{2}\right)}^k{\left(\frac{2}{3}\right)}^{27-k}$ $=\Sigma 27{}^{26}{C}_{k-1}{\left(\frac{1}{2}\right)}^{k-1}\left(\frac{1}{2}\right){\left(\frac{2}{3}\right)}^{26-(k-1)}$

$=27.\frac{1}{2}{\sum }_{k=1}^{27}{}^{26}{C}_{k-1}{\left(\frac{1}{2}\right)}^{k-1}{\left(\frac{2}{3}\right)}^{26-(k-1)}$

$=27\left(\frac{1}{2}\right){(\frac{1}{2}+\frac{2}{3})}^{26}$

$=\frac{27}{2}{\left(\frac{7}{6}\right)}^{26}.$