Prove That $\frac { 1 } { 2 } + \frac { 2 } { 4 } + \frac { 3 } { 8 } + \cdots + \frac { n } { 2 ^ { n } } = \frac { 2 ^ { n + 1 } - 2 - n } { 2 ^ { n } }$

question 38

Short Answer

Prove that $\frac { 1 } { 2 } + \frac { 2 } { 4 } + \frac { 3 } { 8 } + \cdots + \frac { n } { 2 ^ { n } } = \frac { 2 ^ { n + 1 } - 2 - n } { 2 ^ { n } }$ for all $n \geq 1$

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