Calculating the sum of f ( x ) = ∑ n = 0 ∞ n ⋅ 2 n ⋅ x n

Question

Calculating the sum of   f  (  x  )  =      ∑          n      =      0              ∞            n    ⋅          2      n        ⋅          x      n

Gilbert Petty · Accepted Answer

∑          n      =      1        ∞    n      t    n    =  t      d          d      t            (          ∑              n        =        1            ∞              t      n        )    =  t      d          d      t            (          t              1        −        t              )    =      t          (      1      −      t              )        2            Now with t=2x we find      ∑          n      =      1        ∞    n  (  2  x      )    n    =            2      x              (      1      −      2      x              )        2

Frances Pham · Accepted Answer

I believe you can sum the series at the point you left off,

\frac{\int f (x) d x}{x} = 1 + 2 x + (2 x)^{2} + (2 x)^{3} + . . . = \frac{1}{1 - 2 x}

So, we work back to f by itself,

\int f (x) d x = \frac{x}{1 - 2 x}

We take derivatives,

f (x) = \frac{(1 - 2 x) - x (- 2)}{(1 - 2 x)^{2}} = \frac{1}{(1 - 2 x)^{2}}

Calculating the sum of f(x)=sum_(n=0)^infty n2^nx^n

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