Evaluting Summation of Mutiplied Terms

[vsector]

Homework Statement

Evaluate the sums: \sum^{n}_{i=1}ia^{i}

Homework Equations

The Attempt at a Solution

I'm assuming that because there's no limit or anything, the professor wants an equation. I know \sum^{n}_{i=1}i is \frac{1}{2}n(n+1) and \sum^{n}_{i=1}a^{i} is \frac{a^{n+1} - 1}{a-1}-1 but how do you deal with them when multiplied? I can't separate it out like I can in addition. Do I use integrals?

 

[grief]

There are different ways of doing this. There is a way to do it using calculus. Think about letting a be a variable, and integrating to get another expression that you know how to evaluate.

 

[vsector]

I'm sorry, I'm really lost.

There's a relation between summation and integration somewhere that I can use to my advantage is what I got from your reply. It's been a while since I've done any Calculus.

Is there a resource available that I can study this more in depth? All I have is an Algorithms textbook and the Appendix is rather sparse.

 

[grief]

*EDIT*

I'm sorry, I realize a different version of the problem is easier to do with calculus. That's the sum of (i+1)a^i.

 

[grief]

Try to evaluate the sum of (i+1)a^i from i=1 to n first. Integrating (i+1)a^i with respect to i, you get a^(i+1). Now, you have n terms in the sum. The integral of the sum is simply the sum of the integrals of the individual terms, i.e. the sum of the terms a^(i+1) from i=1 to n. This you already know a formula for. How would you proceed from here?

 

[grief]

Just to clarify, the method I'm proposing is NOT treating the sum as an integral. Rather, it's to integrate the expression to yield an easier expression which you can evaluate. Then you would need to take the derivative of that to get back the desired result.

I'll show you a different method as well, which doesn't use Calculus.

Let S represent the sum S=a+2a^2+...+na^n. Multiplying by a, we get aS=a^2+2a^3+...na^(n+1). Subtracting the latter equation from the former, we get (1-a)S=a+a^2+a^3+...+a^n-na^(n+1). The right side consists of a summation you already know how to evaluate (a+a^2+...+a^n) plus an extra term. Now it's just a matter of algebra to find S.

 

[vsector]

Just to clarify, the method I'm proposing is NOT treating the sum as an integral. Rather, it's to integrate the expression to yield an easier expression which you can evaluate. Then you would need to take the derivative of that to get back the desired result.

I'll show you a different method as well, which doesn't use Calculus.

Let S represent the sum S=a+2a^2+...+na^n. Multiplying by a, we get aS=a^2+2a^3+...na^(n+1). Subtracting the latter equation from the former, we get (1-a)S=a+a^2+a^3+...+a^n-na^(n+1). The right side consists of a summation you already know how to evaluate (a+a^2+...+a^n) plus an extra term. Now it's just a matter of algebra to find S.

Brilliant!

Lesse,

S = a + 2a^2 + 3a^3 + \ldots + na^n
aS = a^2 + 2a^3 = 3a^4 + \ldots + na^{n+1}

Subtracting, I get:

S(1 - a) = a + a^2 + a^3 + a^4 + \ldots + a^n + na^{n + 1}
S(1 - a) = \frac{a^{n+1} - 1}{a - 1} - 1 + na^{n + 1}
S = \frac{a^{n+1} - 1}{(a - 1)(1 - a)} - \frac{1}{1 - a} + \frac{na^{n + 1}}{1 - a}
S = \frac{a^{n+1} - 1 - (a - 1) + na^{n+1}(a - 1)}{(a - 1)(1 - a)}
S = \frac{a^{n+1} - a + na^{n+1}(a - 1)}{(a - 1)(1 - a)}

I guess I can't get any neater than that.

EDIT: Whoops. Did it incorrectly.

S(1 - a) = \frac{a^{n+1} - 1}{a - 1} - 1 - na^{n + 1}
S = \frac{a^{n+1} - 1}{(a - 1)(1 - a)} - \frac{1}{1 - a} - \frac{na^{n + 1}}{1 - a}
S = \frac{a^{n+1} - 1 - (a - 1) - na^{n+1}(a - 1)}{(a - 1)(1 - a)}
S = \frac{a^{n+1} - a - na^{n+1}(a - 1)}{(a - 1)(1 - a)}
S = \frac{a^{n+1} - a - na^{n+1}(a - 1)}{(a - 1)(1 - a)}

The other way you mentioned:

Erm...

\sum^n_{i=1}\int(i + 1)a^i di = \sum^n_{i=1}a^{i+1} = \frac{a^{n+1} - 1}{a - 1} - 1 - a + a^{n+1}

Then I'm suppose to take the derivative of that right?

But how is (i + 1)a^i related to ia^i?

... every term gets multiplied by +, how is that difference accounted for?

Thank you for your patience.

 

[Char. Limit]

(i+1)a^i = i a^i + a^i

Taking sums, we get something like this:

\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n i a^i + \sum_{i=1}^n a^i

The first term is one that you can figure out, as it's the derivative of a^(i+1), and the derivative of a sum is the sum of the derivatives, so you can represent it as:

\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n \left(\frac{d}{da} a^{i+1}\right) = \frac{d}{da} \left(\sum_{i=1}^n a^{i+1}\right)

The third term is also simple. Using those two, you can find the value of the second term.

 

[Ray Vickson]

There are different ways of doing this. There is a way to do it using calculus. Think about letting a be a variable, and integrating to get another expression that you know how to evaluate.

That is a standard method, but in this case you need to *differentiate*, not integrate. Think of i*a^i as a*(d/da)(a^i).

RGV

 

[vsector]

(i+1)a^i = i a^i + a^i

Taking sums, we get something like this:

\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n i a^i + \sum_{i=1}^n a^i

The first term is one that you can figure out, as it's the derivative of a^(i+1), and the derivative of a sum is the sum of the derivatives, so you can represent it as:

\sum_{i=1}^n (i+1) a^i = \sum_{i=1}^n \left(\frac{d}{da} a^{i+1}\right) = \frac{d}{da} \left(\sum_{i=1}^n a^{i+1}\right)

The third term is also simple. Using those two, you can find the value of the second term.

Oh! I see. So that was what he meant.

Thanks.