Question

${\sum }_{r=1}^n\left({\sum }_{k=0}^{r-1}{}^n{C}_r{}^r{C}_k{2}^k\right)$ is equal to

• A. 4ⁿ – 3ⁿ
• B. 4ⁿ – 3ⁿ + 1
• C. 4ⁿ – 3ⁿ – 1
• D. 4ⁿ – 3ⁿ + 2

Answer 1

The correct option is A

4ⁿ – 3ⁿ

${\sum }_{r=1}^n{}^n{C}_r({\sum }_{k=0}^n{}^r{C}_k{2}^k-2^n)$
${\sum }_{r=1}^n{}^n{C}_r{3}^n-2^n=4^n-3^n$