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rnd
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#1
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A post about IITJEE resurfaced not long ago. I hadn't heard of it so I looked it up and found this.

http://www.iitjee.org/iit-question-p...s-paper-2.html

It's mostly very tough for me but I liked the question about the sum 1+4x+9x^2+16x^3+......

I would have posted this in the maths society thread but it's probably too easy for those guys.
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Unbounded
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#2
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(Original post by rnd)
It's mostly very tough for me but I liked the question about the sum 1+4x+9x^2+16x^3+......
Nice question :smile:
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qgujxj39
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#3
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Indeed, lovely question.

Am I right in saying that the answer is
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\dfrac{1+x}{(1-x)^3}

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Unbounded
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(Original post by tommm)
Am I right in saying that the answer is
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\dfrac{1+x}{(1-x)^3}

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:yep:
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generalebriety
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#5
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(Original post by tommm)
Indeed, lovely question.

Am I right in saying that the answer is
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\dfrac{1+x}{(1-x)^3}

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Yep. How did you (or anyone else) get it?
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qgujxj39
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#6
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(Original post by generalebriety)
Yep. How did you (or anyone else) get it?
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Let S = 1 + 4x + 9x^2 + \dots = \displaystyle\sum^{\infty}_{r=1} r^2x^{r-1}

Then \int S \mathrm{d}x = \displaystyle\sum^{\infty}_{r=1}  rx^r (plus an arbitrary constant which we can ignore)

= x \displaystyle\sum^{\infty}_{r=1}  rx^{r-1}

= x \frac{d}{dx} \displaystyle\sum^{\infty}_{r=1}  x^r

Notice the sum in the expression above is a geometric series, and we get

\int S \mathrm{d}x = x \frac{d}{dx}\frac{x}{1-x}

Compute the derivative on the RHS, then differentiate both sides and the answer will come out after a bit of manipulation.
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generalebriety
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(Original post by tommm)
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Then \int S \mathrm{d}x = \displaystyle\sum^{\infty}_{r=1}  rx^r (plus an arbitrary constant which we can ignore)
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Just a picky point, but I wouldn't say we can "ignore" an arbitrary constant. What you're really doing is \int_0^x S(t) \mbox{d}t.


(Original post by tommm)
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= x \displaystyle\sum^{\infty}_{r=1}  rx^{r-1}

= x \frac{d}{dx} \displaystyle\sum^{\infty}_{r=1}  x^r
Oh, clever. I didn't do this.

Challenge to anyone who did it vaguely like this:
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Can you do it without getting calculus involved?
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qgujxj39
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#8
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(Original post by generalebriety)
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Just a picky point, but I wouldn't say we can "ignore" an arbitrary constant. What you're really doing is \int_0^x S(t) \mbox{d}t.
Yeah, that's probably a better thing to do. I actually wrote out +C in all my lines of working, and it disappears when we differentiate both sides at the end.
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Unbounded
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#9
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(Original post by generalebriety)
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Can you do it without getting calculus involved?
Would this work?
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We can prove that the generating function for the triangular numbers is essentially  \dfrac{1}{(1-x)^3} as

 (1-x)^{-3} = 1 + 3x + 6x^2 + \cdots + a_n x^n + \cdots

Where  a_n = \dfrac{3 \times 4 \times \cdots \times (n+2)}{n!}

 \iff a_n = \dfrac{(n+1)(n+2)}{2} = T_{n+1} , where  T_n = \dfrac{n(n+1)}{2} is the nth triangular number.

And noting that  T_n + T_{n+1} = (n+1)^2 for all n, then it follows that

 (1+x)(1-x)^{-3} = \displaystyle\sum_{n=0}^{\infty} (T_n + T_{n+1})x^n

 \therefore \dfrac{1+x}{(1-x)^3} = 1 + 4x + 9x^2 + \cdots + (n+1)^2x^n  + \cdots
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generalebriety
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#10
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(Original post by GHOSH-5)
Would this work?
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We can prove that the generating function for the triangular numbers is essentially  \dfrac{1}{(1-x)^3} as

 (1-x)^{-3} = 1 + 3x + 6x^2 + \cdots + a_n x_n + \cdots

Where  a_n = \dfrac{3 \times 4 \times \cdots \times (n+2)}{n!}

 \iff a_n = \dfrac{(n+1)(n+2)}{2} = T_{n+1}

And noting that  T_n + T_{n+1} = (n+1)^2 for all n, then it follows that

 (1+x)(1-x)^{-3} = 1 + 2^2x + 3^2x^2 + \cdots = 1 + 4x + 9x^2 + \cdots + (n+1)^2x^n  + \cdots
Nicely done.

I went for something a bit messier but a bit less plucked out of thin air. This is all still a bit 'raw', as I've tried to explain my train of thought along the way:

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Notice that:
1 + 4x + 9x^2 + 16x^3 + ...
= 1 + 3x + 5x^2 + 7x^3 + ...
+ x + 3x^2 + 5x^3 + ...
+ x^2 + 3x^3 + ...
+ x^3 + ...
+ ...

i.e. we're getting the pattern "1, 3, 5, 7, 9, 11, ..." on each line, which can be made nicely rigorous by proving (by induction) that \displaystyle\sum_{k=1}^n (2k-1) = n^2.

So we can rewrite our series as \displaystyle \sum_{n=1}^\infty \sum_{k=1}^n (2k-1)x^{n-1}, and swap the sums to get \displaystyle \sum_{k=1}^\infty \sum_{n=k}^\infty (2k-1)x^{n-1}, and the inner sum is just a GP, so that's easy to evaluate. (If swapping the sums is too much like hard work, just look at the way I split the series up before, and instead of reading across, read down, so that you get

1 + x + x^2 + x^3 + ...
+ 3x + 3x^2 + 3x^3 + ...
+ 5x^2 + 5x^3 + ...
+ 7x^3 + ...
+ ...

which is exactly what this new sum is, and it's very easy to sum the rows now as they're all GPs.)

Anyway, clear all that algebra up and you end up having to sum \sum_1^\infty kx^{k-1}, which is just \sum_1^\infty x^{k-1} + x\sum_1^\infty (k-1)x^{k-2}. But \sum_1^\infty kx^{k-1} = \sum_1^\infty (k-1)x^{k-2} (as the first term on the RHS is 0, and that's the only place they differ). Put it all together, and you get a messy but calculus-free solution.
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Unbounded
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#11
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(Original post by generalebriety)
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Notice that:
1 + 4x + 9x^2 + 16x^3 + ...
= 1 + 3x + 5x^2 + 7x^3 + ...
+ x + 3x^2 + 5x^3 + ...
+ x^2 + 3x^3 + ...
+ x^3 + ...
+ ...
Nice :smile:
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rnd
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#12
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My way was as follows.
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S=1+4x+9x^2+.....

Sx=x+4x^2+9x^3....

S-Sx=1+3x+5x^2+7x^3+.......

(S-Sx)x=x+3x^2+5x^3+.....

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

S(1-x)^2=(1+x)/(1-x)

S=(1+x)/(1-x)^3

I'm sure I've seen something like this before, otherwise I probably wouldn't have solved it.
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Unbounded
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#13
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(Original post by rnd)
My way was as follows.
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S=1+4x+9x^2+.....

Sx=x+4x^2+9x^3....

S-Sx=1+3x+5x^2+7x^3+.......

(S-Sx)x=x+3x^2+5x^3+.....

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

S(1-x)^2=(1+x)/(1-x)

S=(1+x)/(1-x)^3
That's very neat
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around
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#14
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Calculus all the way baby.

elementary methods are for luz0rs (<- safety smiley)
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AsakuraMinamiFan
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#15
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Anybody else having trouble with some characters not showing on the page? (I'm using Fx on Linux.) Is there a specific page encoding it should have? Thanks.
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Totally Tom
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#16
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(Original post by AsakuraMinamiFan)
Anybody else having trouble with some characters not showing on the page? (I'm using Fx on Linux.) Is there a specific page encoding it should have? Thanks.
same. you can guess most of them though.
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Simplicity
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#17
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Can anyone prove it by induction?

Then prove induction works?

P.S. Apart from calculus is there a more advanced way to solve it, for example fourier series?
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generalebriety
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#18
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(Original post by Simplicity)
Can anyone prove it by induction?

Then prove induction works?

P.S. Apart from calculus is there a more advanced way to solve it, for example fourier series?
I hope you're joking.
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Totally Tom
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#19
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(Original post by Simplicity)
Can anyone prove it by induction?

Then prove induction works?

P.S. Apart from calculus is there a more advanced way to solve it, for example fourier series?
You're such an Andrew Ewart it's unbelievable.
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qgujxj39
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#20
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Fourier series? From the basic grounding I have in Fourier series, I can see no connection whatsoever between them and this question.

And how would induction work when you're proving a proposition about x, which isn't necessarily integral?
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