the summation from 0 to infinity of p*x^(p-1) for -1
Please show steps.
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As p goes to infinity, we have:
1 + 2x + 3x^2 + 4x^3 + .. + p*x^(p-1).
Notice that taking the integral of the series yields:
oo
Σ x^p = x + x^2 + x^3 + x^3 + . . . + x^p.
p=0
This geometric series has a ratio of x. Therefore we can use:
a/(1 - r), where a = first term and r = ratio. Doing so, yields:
F(x) = x/(1 - x).
Differentiating this function gives:
f(x) = 1/(x - 1)^2.
Therefore, the series representation of p*x^(p-1) is:
oo
Σ 1/(1 - x)^2
n=1
Hope this helped.
1 + 2x + 3x^2 + 4x^3 + .. + p*x^(p-1).
Notice that taking the integral of the series yields:
oo
Σ x^p = x + x^2 + x^3 + x^3 + . . . + x^p.
p=0
This geometric series has a ratio of x. Therefore we can use:
a/(1 - r), where a = first term and r = ratio. Doing so, yields:
F(x) = x/(1 - x).
Differentiating this function gives:
f(x) = 1/(x - 1)^2.
Therefore, the series representation of p*x^(p-1) is:
oo
Σ 1/(1 - x)^2
n=1
Hope this helped.
-
Starting from p=1
1+2x+3x^2 +4x^3+ n*x^(n-1)++++
the integral is C+x+x^2+x^3++++++ x^n + which is a geometric series with r=x
so the sum is S(x)= C+ x/1-x
Taking the derivative of the function
f(x) =1/(x-1)^2 your sum
1+2x+3x^2 +4x^3+ n*x^(n-1)++++
the integral is C+x+x^2+x^3++++++ x^n + which is a geometric series with r=x
so the sum is S(x)= C+ x/1-x
Taking the derivative of the function
f(x) =1/(x-1)^2 your sum