Question:
The sum of the series $2 \cdot{ }^{20} \mathrm{C}_{0}+5 \cdot{ }^{20} \mathrm{C}_{1}+8 \cdot{ }^{20} \mathrm{C}_{2}+11 \cdot{ }^{.20} \mathrm{C}_{3}$ $+\ldots+62 \cdot{ }^{20} \mathrm{C}_{20}$ is equal to :
Correct Option: , 2
Solution:
$2 \cdot{ }^{20} \mathrm{C}_{0}+5 \cdot{ }^{20} \mathrm{C}_{1}+8 \cdot{ }^{20} \mathrm{C}_{2}+\ldots \ldots+62 \cdot{ }^{20} \mathrm{C}_{20}$
$=\sum_{r=0}^{20}(3 r+2){ }^{20} C_{r}=3 \sum_{r=0}^{20} r \cdot{ }^{20} C_{r}+2 \sum_{r=0}^{20}{ }^{20} C_{r}$
$=60 \sum_{r=1}^{20}{ }^{19} C_{n-1}+2 \sum_{r=0}^{20}{ }^{20} C_{r}$
$=60 \times 2^{19}+2 \times 2^{20}=2^{21}[15+1]=2^{25}$