Question:
{C0 + 2C1 + 3C2 + …+ (n + 1)Cn} = ?
{C0 + 2C1 + 3C2 + …+ (n + 1)Cn} = ?
Options:
- (n + 1) · 2n
- (n + 2) · 2n-1
- n · 2n-1
- (n + 2) · 2n+1
- (n + 1) · 2n
- (n + 2) · 2n-1
- n · 2n-1
- (n + 2) · 2n+1
Correct Answer: b)
Explanation:
{C0 + 2C1 + 3C2 + …+ (n + 1)Cn} = (n + 2) · 2n-1
{C0 + 2C1 + 3C2 + …+ (n + 1)Cn} = (n + 2) · 2n-1