Question

The sum of the series $2.{}^{20}{C}_0+5.{}^{20}{C}_1+8.{}^{20}{C}_2+11.{}^{20}{C}_3+....+62.{}^{20}{C}_{20}$ is equal to :

• A. $2^{24}$
• B. $2^{25}$
• C. $2^{26}$
• D. $2^{23}$

Answer 1

Answer: (B)

$2{.}^{20}{C}_0+5{.}^{20}{C}_1+8{.}^{20}{C}_2+11{.}^{20}{C}_3+....+62{.}^{20}{C}_{20}$
$={\sum }_{r=0}^{20}{\left(3r+2\right)}^{20}{C}_r$
$=3{\sum }_{r=0}^{20}.{}^{20}{C}_r+2{\sum }_{r=0}^{20}{}^{20}{C}_r$
$=3{\sum }_{r=0}^{20}r{\left(\frac{20}{r}\right)}^{19}{C}_{r-1}+2.2^{20}$
$=60.2^{19}+2.2^{20}=2^{25}$