Summing i*r^(i-1) from i=1 to i=n

[mathvision]

I wanted to post this in the homework forum, but there's only pre-calc for math.

Question: Show that i*r^i-(i-1)*r^(i-1) = r^(i-1)-(1-r)i*r^i-1. Use this result to find the sum of i*r^(i-1) from i=1 to i=n.

I've done the first part of this question, but need some help with the second. Thanks!

 

[MathematicalPhysicist]

This identity isn't valid for i=1, check it.

 

[HallsofIvy]

I wanted to post this in the homework forum, but there's only pre-calc for math.

Question: Show that i*r^i-(i-1)*r^(i-1) = r^(i-1)-(1-r)i*r^i-1. Use this result to find the sum of i*r^(i-1) from i=1 to i=n.

I've done the first part of this question, but need some help with the second. Thanks!

Do you mean "i*r^i-(i-1)*r^(i-1) = r^(i-1)-(1-r)i*r^(i-1)"? Otherwise, as MathematicalPhysicist said, it is not true for i= 1.

In any case, I would not use that. Are you required to?

I would note that \sum_{i=1}^n r^i is a geometric sum and that \sum_{i= 1}^n i r^{i-1} is its derivative with respect to r.

 

[mathvision]

Do you mean "i*r^i-(i-1)*r^(i-1) = r^(i-1)-(1-r)i*r^(i-1)"? Otherwise, as MathematicalPhysicist said, it is not true for i= 1.

In any case, I would not use that. Are you required to?

I would note that \sum_{i=1}^n r^i is a geometric sum and that \sum_{i= 1}^n i r^{i-1} is its derivative with respect to r.

Yes it was a typo. And yes I'm required to use that result!

 

[uart]

I wanted to post this in the homework forum, but there's only pre-calc for math.

Question: Show that
i r^i - (i-1)r^{i-1} = r^{i-1} - (1-r)i r^{i-1}

Use this result to find the sum of i r^{i-1} from i=1 to i=n.

I've done the first part of this question, but need some help with the second. Thanks!

Hi mathvision, I've edited that for you.

If this isn't pre-calc then just differentiating the geometric series as Halls suggests is the best way to do it.

If this is pre-calc and you need to do it without using calculus then you can do the following. Sum both sides of your above expression and notice that the RHS is the sum of (r-1) times the sum you are trying to find, plus a simple geometric series. Also, the left hand side completely "telescopes", leaving just the first and last terms. Basically then you can just rearrange this to get an expression for the part you are interested in.