How would you determine the power series of 1/(1-x)^3. I know that the series of 1/(1-x) is x^n, but how would you manipulate it for this scenario?

I got 1+3x+6x^2+10x^3+15x^4... but I was wondering how I would write a series for this.

surely you recognize the sequence

1, 3, 6, 10, ... as n(n+1)/2

it is the sequence of partial sums of

1+2+3+4+...

To determine the power series of 1/(1-x)^3, you can start by understanding the relationship between power series and differentiation. The power series expansion of a function can often be obtained by differentiating known power series expansions.

Step 1: Start with the known power series expansion of 1/(1-x):
1/(1-x) = 1 + x + x^2 + x^3 + ...

Step 2: Differentiate both sides of the equation:
(d/dx) (1/(1-x)) = (d/dx)(1 + x + x^2 + x^3 + ...)
(d/dx) (1/(1-x)) = 0 + 1 + 2x + 3x^2 + ...

Step 3: Apply the power rule of differentiation to simplify the expression on the right:
(d/dx) (1/(1-x)) = 1/(1-x)^2

Step 4: Multiply both sides by (1-x)^2 to isolate 1 on the right-hand side:
(1/(1-x))^2 = 1 + 2x + 3x^2 + ...

Step 5: Finally, differentiate both sides again:
(d/dx) [(1/(1-x))^2] = (d/dx)(1 + 2x + 3x^2 + ...)
(d/dx) [(1/(1-x))^2] = 0 + 2 + 2(3)x + 3(2)x^2 + ...

Step 6: Apply the power rule of differentiation again to simplify the expression on the right:
(d/dx) [(1/(1-x))^2] = 2/(1-x)^3

Hence, we have obtained the power series expansion for 1/(1-x)^3:

2/(1-x)^3 = 2 + 6x + 12x^2 + ...

Therefore, the power series expansion of 1/(1-x)^3 is 2 + 6x + 12x^2 + ...