Infinite sum problem 1994 SGS (1 Viewer)

[asianese]

7b)

 

[deswa1]

No your third line is wrong. You can only sum to infinity like that if -1<x<1. The question doesn't ask for infinity anyway, just up to n.

 

[deswa1]

Don't know why my post isn't coming up- What I'm trying to say is you can only sum to inifinity like that if the absolute value of x is less than one and the question doesn't even ask for a summation to infinity, just n.

 

[asianese]

Oh yeah O_O whoa completely missed that

 

[deswa1]

Oh yeah O_O whoa completely missed that

No worries- its a relatively common mistake. An easy way to check something like that though is sub in some massive value like x=100. The LHS becomes huge but the RHS becomes tiny-> so you can easily conclude you made a mistake. For questions like this, always sub in something big though so the effect is even more obvious and you can confirm without a calc.

 

[asianese]

i)

is all i got now

 

[deswa1]

i)

is all i got now

Yep thats right so far. Now start a new G.P with x^2 as the first term, then another with x^3 as the first term etc. and sum them. They will all end on the same denominator and simplify nicely.

 

[Godmode]

Alrighty, brace for a complete mess; I have no idea how to use LaTex.

let S = x + 2x^2 + 3x^3 + ... + nx^n

now, xS = x^2 + 2x^3 + 3x^4 + ... + (n - 1)x^n + nx^(n+1)

subtracting xS from S, we get

S - xS = x + 2x^2 + 3x^3 + 4x^4 + ... + nx^n - (x ^2 + 2x^3 + 3x^4 + ... + (n - 1)x^n) - nx^(n+1)

after collecting like terms, we get

(1 - x)S = x + x^2 + x^3 + ... + x^n - nx^(n+1)

using the result from i. ,

(1 - x)S = x(x^n - 1)/(x -1) - nx^(n+1)

dividing through by (1 - x)

so S = x^(n + 1) - x/(x -1)(1 - x) - nx^(n+1)/(1 - x)

flipping the sign of the (1 -x) in the denominator,

S = nx^(n+1)/(x-1) - x^(n + 1)/(x - 1)^2 + x/(x - 1)^2
= [ n(x-1)x^(n+1) - x^(n+1) + x ]/(x - 1)^2
= [ nx^(n+2) - (n+1)x^(n+1) + x ] /(x-1)^2

therefore, x + 2x^2 + 3x^3 + ... + nx^n = [nx^(n + 2) - (n + 1)x^(n+1) + x]/(x-1)^2

just check my working, i may have made a mistake lol.

 

[asianese]

Ahhh Yes.

For some reason I had all the separate sums, on the denominator 1-x, 1-x^2, 1-x^3 etc instead of just 1-x 1-x 1-x...

Thanks the both of you! And I checked it Godmode it seems good

 

[Godmode]

No worries lol; one of these days, I swear I will learn LaTex.

Now I was wondering about the actual series in question, do we actually get the same result by summing each G.P. together? Or do we get another expression, and does it also tend to infinity? If they are different, which one 'grows' faster?

 

[seanieg89]

By the way, this question can also be done by differentiation of the geometric series.

 

[seanieg89]

(The bottom expression is x times the derivative of the top one.)

 

[Godmode]

Haha I was thinking of that. That's how I got the idea behind my solution.

 

[zhiying]

Here's an alternative method that splits the sum that we cant apply AP/GP formulas for into n other sums that we can use those formulas


Edit: woo got the same answer as Godmode so I think it's right

 

[zhiying]

Also Godmode if you cbf learning LaTex, or don't know how to (like me) you can get MathType and do slightly less quality stuff on MS word

 

[asianese]

Btw Zhiying nice work on 99.75! I'm guessing Jap didn't count? haha.. Do you think it was hard getting 93? The class last year (your year) only got 1 Band 6

 

[zhiying]

Yeah I did Japanese only because I liked it, it was pretty easy for me. Most of my class just didn't study much. The year above me had no band 6

 

[asianese]

Oh ok - A little more confident then lol.

A girl in my class is likely to state rank O_O

 

[RealiseNothing]

A pretty cool way:





Now notice that the bit inside the bracket is the derivative of the first expression, so we find the derivative of the closed sum of the first expression:



Simplifying gives:



Now we still have that 'x' out the front of the brackets from the second line, so lets multiply that in:



Simplifying gives:



Which is the answer.

 

[zhiying]

Who? And what year are you in