Sum of 1 + 2a + 3a2 + ... to n terms

[lionely]

Homework Statement

Sum the series 1 + 2a + 3a² + ... to n terms



This series consists of an a.p. (with general term n) and gp general term a^(n-1)

right?

So the series general term is na^(n-1)

So is the sum the sum of each progression times each other?

i.e (1-a^n)/(1-a) * (n(n+1)/2) ?

 

[LCKurtz]

Homework Statement

Sum the series 1 + 2a + 3a² + ... to n terms



This series consists of an a.p. (with general term n) and gp general term a^(n-1)

right?

So the series general term is na^(n-1)

So is the sum the sum of each progression times each other?

i.e (1-a^n)/(1-a) * (n(n+1)/2) ?

No. Hint: What is the derivative of $x + x^2 + x^3+ x^4 ...$?

 

[lionely]

The derivative would be 1 + 2x + 3x^2 ..., so are you saying I should integrate that series? to make it into a g.p with common ratio umm a?

 

[LCKurtz]

I'm suggesting you think about how you can use that fact to work your problem.

 

[Curious3141]

Another way, longer, but that doesn't require calculus, is to consider the sum of the following geometric series:

$\displaystyle S_1 = 1 + a + a^2 + ... a^{n-1}$

$\displaystyle S_2 = ~~~~~~~a + a^2 + ... a^{n-1}$

$\displaystyle S_3 = ~~~~~~~~~~~~~~a^2 + ... a^{n-1}$

$\displaystyle ...$

$\displaystyle S_{n} = ~~~~~~~~~~~~~~~~~~~~~~~~~a^{n-1}$

 

[lionely]

Well I got this

Sn = 1 + 2a + 3a^2 + 4a^3 ...+ (n-1)a^(n-2) + na^(n-1)

aSn = a + 2a^2 + 3a^3 ...+ (n-1)a^(n-1) + na^n

Subtracting the sums

Sn(1-a) = 1 + a + a^2 + a^3 ...+ a^(n-1) - na^n

Sn(1-a) = 1 + (1-a^n)/(1-a) - na^n ( summed the G.P.)

Sn = (1-a^n)/(1-a) + (1-a^n)/(1-a)^2Is this like what you wanted me to do with the differential sum? and Curious I don't really get your way :(

 

[Curious3141]

Well I got this

Sn = 1 + 2a + 3a^2 + 4a^3 ...+ (n-1)a^(n-2) + na^(n-1)

aSn = a + 2a^2 + 3a^3 ...+ (n-1)a^(n-1) + na^n

Subtracting the sums

Sn(1-a) = 1 + a + a^2 + a^3 ...+ a^(n-1) - na^n

Sn(1-a) = 1 + (1-a^n)/(1-a) - na^n ( summed the G.P.)

Sn = (1-a^n)/(1-a) + (1-a^n)/(1-a)^2


Is this like what you wanted me to do with the differential sum? and Curious I don't really get your way :(

EDIT: There is an error in your working. Check your geometric sum in the second last step.

Yes, this is a neat way to do it so good on you. I've done it your way before, but for some reason, it escaped my brain today.

The way I suggested in my post would've been much longer, but basically each of those series are geometric sums. Add them up and you get the series in question. If you do the algebra and add up the expressions for each geometric sum, they will simplify.

LCKurtz's suggestion (I think) was to get a compact expression for the geometric sum, then differentiate it wrt a.

Three ways, which will all give what you got.

BTW, you can simplify your expression a little (once you correct your error). Try to get everything over a common denominator $(a-1)^2$.

 

[lionely]

This method i used should work if I try to sum

1 + 4(1/2) + 7(1/2)^2 + ... 28(1/2)^9 right?

I keep trying but I keep getting the wrong answer!

 

[lionely]

Never mind it worked ! I'm just an idiot.

 

[Curious3141]

Never mind it worked ! I'm just an idiot.

Yes, I hope you corrected the error in your expression. A good way to test a general formula like this is to put in a value to 3 terms and try it out. But don't use a "nice value", like 2, 3 or even 1/2 or whatever. The safest way is to use something like $\pi$. This is what I did to test my answer. When you put in a transcendental number there is no way to get the sum working out just so by "accident".

 

[lionely]

Oh okay thanks for the advice!

 

[LCKurtz]

Is this like what you wanted me to do with the differential sum? and Curious I don't really get your way :(

LCKurtz's suggestion (I think) was to get a compact expression for the geometric sum, then differentiate it wrt a.

Yes. $a + a^2 + a^3 +...+a^n = \frac{a-a^{n+1}}{1-a}$. Then differentiate both sides.