# Calculate 1+2x+3x^2+4x^3+---+99x^100.

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### 2 Answers

I assume that the last term of requested summation is `100x^99` , since it needs to follow the pattern of previous terms.

You need to evaluate the given summation such that:

`1+2x+3x^2+4x^3+...+100x^99 = sum_(k=1)^100 k*x^(k-1)`

You need to notice that differentiating the function `y = x^k` with respect to x, yields the product `(dy)/(dx) = k*x^(k-1)` .

`1+2x+3x^2+4x^3+...+100x^99 = (d (sum_(k=1)^100 x^k))/(dx) `

Evaluating the summation `sum_(k=1)^100 x^k` yields:

`sum_(k=1)^100 x^k = (1 - x^101)/(1 - x)`

`1+2x+3x^2+4x^3+---+100x^99 = (d((1 - x^101)/(1 - x)))/(dx)`

Hence, you need to use the quotient rule to evaluate the derivative `(d((1 - x^101)/(1 - x)))/(dx)` , such that:

`(d((1 - x^101)/(1 - x)))/(dx) = ((1 - x^101)'(1 - x) - (1 - x^101)(1 - x))/((1 - x)^2)`

`(d((1 - x^101)/(1 - x)))/(dx) = (-101x^100(1 - x) + 1 - x^101)/((1 - x)^2)`

`(d((1 - x^101)/(1 - x)))/(dx) = (-101x^100 + 101x^101 + 1 - x^101)/((1 - x)^2)`

`(d((1 - x^101)/(1 - x)))/(dx) = (100x^101 - 101x^100 + 1)/((1 - x)^2)`

**Hence, evaluating the summation `1+2x+3x^2+4x^3+...+100x^99` , using derivatives and summation of terms of geometric series, yields **`1+2x+3x^2+4x^3+...+100x^99 = (100x^101 - 101x^100 + 1)/((1 - x)^2).`

**Sources:**

`S=1+2x+3x^2+4x^3+........................+100x^99`

`xS=... +x+2x^2+3x^3+.............................+99x^100+100x^100`

`(1-x)S=1+x+x^2+......+x^99-100x^100`

`(1-x)S=(x^100-1)/(x-1)-100x^100`

`S=(1-x^100)/(1-x)^2-(100x^100)/(1-x)`

This is an example of arithmetico Geometric series.

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Let S be the sum of the given series.

`S=1+2x+3x^2+.............+100x^99` (i)

multiply S by x

`xS=x+2x^2+3x^3+............+99x^99+100x^100` (ii)

subtract (ii) from (i)

`(1-x)S=1+x+x^2+x^3+.....+x^99-100x^100`

`=(1-x^100)/(1-x)-100x^100`

`S=(1-x^100)/(1-x)^2-(100x^100)/(1-x)`

This seris is known as arithmetico geometric series . x is common ratio of the corresponding geometric serie .