Antoun J. Yaacoub Ph.D.

Compute the sum of the following series: \( \sum_{i=1}^n i \times 2^i\).

Difficulty level

This exercise is mostly suitable for students

Suppose S = Sum i=1 to n i . 2^i.
Then S can be written as: 1.2^1 + 2.2^2 + 3.2^3 + ... + n.2^n.
Such summations can often be simplified and solved using "term sliding".
         S  = 1.2   + 2.2^2 + 3.2^3 + ...      + n.2^n
That is, 2S = 1.2^2 + 2.2^3 + 3.2^4 + ... + (n-1).2^n + n.2^(n+1)
Subtract the second equation from the first, and we obtain:
- S = 1.2 + 1.2^2 + 1.2^3 + ... + 1.2^n - n.2^(n+1)
    = 2 + 2^2 + 2^3 + ... + 2^n - n.2^(n+1)
    = 2(1 + 2^1 + 2^2 + ... + 2^(n-1)) - n.2^(n+1)             
By using geometric progression, we obtain :
- S = 2.(2^n-1) - n.2^(n+1)
Therefore, S = (n-1)2^(n+1) + 2

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