Solving summation 2 n + 2 2 ( n − 1 ) + 2 3 ( n − 2 ) + . . . . + 2 n

Question

Solving summation   2  n  +      2    2    (  n  −  1  )  +      2    3    (  n  −  2  )  +  .  .  .  .  +      2    n

boursecasa2je · Accepted Answer

All you have to find out is how to sum out

\sum_{k = 1}^{n} 2^{k} (n - k + 1)

This unwinds as

(n + 1) \sum_{k = 1}^{n} 2^{k} - \sum_{k = 1}^{n} k 2^{k}

The first term is a geometric series. Can you find out what the second term is?
One way is to note that

\sum_{k = 1}^{n} x^{k} = x \frac{1 - x^{n}}{1 - x}

Now, differentiate and multiply by x to get that

\sum_{k = 1}^{n} k x^{k} = x \cdot \frac{d}{d x} (x \frac{1 - x^{n}}{1 - x})

After finding out what the right hand side is, plug x=2.

Jenny Roberson · Accepted Answer

Define      s    n    =      ∑          i      =      0              n      −      1            2          i      +      1        (  n  −  i  )and note that you have the relation:      S          n      +      1        =      S    n    +  2  +      2    2    +  ⋯  +      2          n      +      1        =      S    n    +      2          n      +      2        −  2this comes from the fact that      2          i        (  n  −  i  )  =      2          i        (  n  −  i  +  1  −  1  )  =      2          i        (  n  −  i  −  1  )  +      2    i  Noting that :       S    1    =  2 and that the solution of the simple recurrence above is:      S          n      +      1        =      S    1    +      2    3    +      ⋯    +        2          n      +      2        −  2  n  =      2          n      +      3        −      2    3    +  2  −  2  nThus      S          n        =      2          n      +      2        −      2    3    +  2  −  2  (  n  −  1  )

Solving summation 2n+2^2(n-1)+2^3(n-2)+...+2^n

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