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After spending a good 15 minutes trying to write out my working through the problem without knowing latex, I've given up. I'm trying to prove that the sum of r squared for all integer r between 1 and n is equal to 1/6n(n+1)(2n+1) using calculus.

If anyone is willing to help, could you drop me a pm with an msn address or your preferred IRC server, I think I could explain it in real time chat, but to do so on a forum would probably take an unwieldy amount of follow up posts or corerctions.

If anyone is willing to help, could you drop me a pm with an msn address or your preferred IRC server, I think I could explain it in real time chat, but to do so on a forum would probably take an unwieldy amount of follow up posts or corerctions.

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DoMakeSayThink

After spending a good 15 minutes trying to write out my working through the problem without knowing latex, I've given up. I'm trying to prove that the sum of r squared for all integer r between 1 and n is equal to 1/6n(n+1)(2n+1) using calculus.

If anyone is willing to help, could you drop me a pm with an msn address or your preferred IRC server, I think I could explain it in real time chat, but to do so on a forum would probably take an unwieldy amount of follow up posts or corerctions.

If anyone is willing to help, could you drop me a pm with an msn address or your preferred IRC server, I think I could explain it in real time chat, but to do so on a forum would probably take an unwieldy amount of follow up posts or corerctions.

Calculus!? Why not use induction?

I've done the proof by induction, I just don't like it for some reason. It seems less elegant, maybe? Also, you need to know the formula to prove it, where as with calculus your derive it while proving it.

I think proof by induction is really nice. Anyway, sorry, I'm no good at that stuff with calculus, so I can't help you.

something along the lines of:

Sum to n terms of [(r+1)^3 - r^3] = (n+1)^3 - 1

= Sum to n terms of [3r^2 + 3r + 1]

= Sum to n terms of [3r^2] + Sum to n terms of [3r] + n

= 3* Sum to n terms of [r^2] + 3/2 n(n+1) + n = (n+1)^3 - 1

3* Sum to n terms of [r^2] = (n+1)^3 - 1 - 3/2 n(n+1) - n

3* Sum to n terms of [r^2] = n^3 + 3n^2 + 3n + 1 - 1 - 3/2 n^2 - 3/2n - n

3* Sum to n terms of [r^2] = n^3 + 3/2n^2 + 1/2n

Sum to n terms of [r^2] = 1/6 [2n^3 + 3n^2 + n] = 1/6n (2n^2 + 3n + 1)

Sum to n terms of [r^2] = 1/6 n(2n+1)(n+1)

Voila!

Sum to n terms of [(r+1)^3 - r^3] = (n+1)^3 - 1

= Sum to n terms of [3r^2 + 3r + 1]

= Sum to n terms of [3r^2] + Sum to n terms of [3r] + n

= 3* Sum to n terms of [r^2] + 3/2 n(n+1) + n = (n+1)^3 - 1

3* Sum to n terms of [r^2] = (n+1)^3 - 1 - 3/2 n(n+1) - n

3* Sum to n terms of [r^2] = n^3 + 3n^2 + 3n + 1 - 1 - 3/2 n^2 - 3/2n - n

3* Sum to n terms of [r^2] = n^3 + 3/2n^2 + 1/2n

Sum to n terms of [r^2] = 1/6 [2n^3 + 3n^2 + n] = 1/6n (2n^2 + 3n + 1)

Sum to n terms of [r^2] = 1/6 n(2n+1)(n+1)

Voila!

That's a nice proof, but it's still not really what I was thinking. I'll write it out in a somewhat presentable format another time, but it's 10 to 3 in the morning and I've got college tomorrow. BLERGH.

Calculus is for continuous variables. You can't really use it for summations.

If you do, you will have to use a continuity correction or find the right integration constant. (And I'm not sure even that work work as a rigorous proof). I'm not sure about that.

If you just want to find out how the original formula was found, you can do this: (OCR FP1 textbook)

- Assume that sum of r^2 is given by a cubic polynomial (since sum of r is given by a quadratic, and sum of r^0 is given by a linear)

- Calculate sum of r^2 with n=1,2 and 3

- Form simultaneous equations and solve.

- Now you've shown that if sum of r^2 is a cubic polynomial, then it's the polynomial you reached.

- Prove the result by induction or the method of differences (shown above)

If you do, you will have to use a continuity correction or find the right integration constant. (And I'm not sure even that work work as a rigorous proof). I'm not sure about that.

If you just want to find out how the original formula was found, you can do this: (OCR FP1 textbook)

- Assume that sum of r^2 is given by a cubic polynomial (since sum of r is given by a quadratic, and sum of r^0 is given by a linear)

- Calculate sum of r^2 with n=1,2 and 3

- Form simultaneous equations and solve.

- Now you've shown that if sum of r^2 is a cubic polynomial, then it's the polynomial you reached.

- Prove the result by induction or the method of differences (shown above)

Another quickish way that I just dreamed up is similar to r3mot's. First assume that you can express the sum as a cubic:

$f(n) \equiv \sum_{r=1}^n r^2 = a_0 + a_1n + a_2n^2 + a_3n^3$

Then notice that

$f(n)-f(n-1)=n^2$

and so

$(a_1-a_2+a_3) + (2a_2-3a_3)n + 3a_3n^2 = n^2$

From this you can quickly deduce that $a_3=1/3$, and thus $a_2=1/2$ and $a_1=1/6$. To get the value for $a_0$ you simply realise that

$f(1)=a_0+a_1+a_2+a_3=1$

and thus you must have $a_0=0$. Then as r3mot says, you've shown that IF it is a cubic, then it is the cubic you've found. Then go on to show that your cubic is the correct expression.

$f(n) \equiv \sum_{r=1}^n r^2 = a_0 + a_1n + a_2n^2 + a_3n^3$

Then notice that

$f(n)-f(n-1)=n^2$

and so

$(a_1-a_2+a_3) + (2a_2-3a_3)n + 3a_3n^2 = n^2$

From this you can quickly deduce that $a_3=1/3$, and thus $a_2=1/2$ and $a_1=1/6$. To get the value for $a_0$ you simply realise that

$f(1)=a_0+a_1+a_2+a_3=1$

and thus you must have $a_0=0$. Then as r3mot says, you've shown that IF it is a cubic, then it is the cubic you've found. Then go on to show that your cubic is the correct expression.

Okay, I'll attempt to write out my working. I've showed it to several maths graduates, and they seem to think it's a dead end, so I'm inclined to agree, but just incase there are any sparks of inspiration, I'll do my best to write out in LaTeX what I've got.

$\sum r^2 = \sum \displaystyle\int \frac{d}{dr} (r^2) dr = \displaystyle\int \sum (2r) \quad dr = 2 \displaystyle\int \sum r \ dr$

We already have a proof for $\sum r = \frac{1}{2}n(n+1)$

So $2 \displaystyle\int \sum r \ dr = 2 \displaystyle\int \frac{1}{2}n(n+1)\ dr = \displaystyle\int n^2 + n \ dr$

And here, we get stuck, because it's integrating with respect to r, and there is no relation between r and n, however, if we could somehow get it to differentiate with respect to n, I *think* it would come out as:

$\frac{1}{3}n^3 + \frac{1}{2}n^2 + cn$

I think you could express cn as c, because it's a constant, but if you leave it as cn it seems to be that little bit closer the formula we're looking for. The reason it's cn in the first place, is because there would be n constants of intgeration, because you're intgerating $1^2 + 2^2 + 3^2+ ... +n^2$

My apologies if the TeX is rather poorly formatted, this is my first ever usage of it.

$\sum r^2 = \sum \displaystyle\int \frac{d}{dr} (r^2) dr = \displaystyle\int \sum (2r) \quad dr = 2 \displaystyle\int \sum r \ dr$

We already have a proof for $\sum r = \frac{1}{2}n(n+1)$

So $2 \displaystyle\int \sum r \ dr = 2 \displaystyle\int \frac{1}{2}n(n+1)\ dr = \displaystyle\int n^2 + n \ dr$

And here, we get stuck, because it's integrating with respect to r, and there is no relation between r and n, however, if we could somehow get it to differentiate with respect to n, I *think* it would come out as:

$\frac{1}{3}n^3 + \frac{1}{2}n^2 + cn$

I think you could express cn as c, because it's a constant, but if you leave it as cn it seems to be that little bit closer the formula we're looking for. The reason it's cn in the first place, is because there would be n constants of intgeration, because you're intgerating $1^2 + 2^2 + 3^2+ ... +n^2$

My apologies if the TeX is rather poorly formatted, this is my first ever usage of it.

What you're doing is saying

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x dx$

but then from here we can't interchange the integration and summation because the integral is dependent on r.

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x dx$

but then from here we can't interchange the integration and summation because the integral is dependent on r.

*sniff* and it could've been so beautiful!

I realise that you get a bit suck, and I'm sure I've seen some strange thing to do with messing about with limits, but I don't think that gets us anywhere anyway. All of the messing about with limits stuff is a tad beyond my scope anyway.

I realise that you get a bit suck, and I'm sure I've seen some strange thing to do with messing about with limits, but I don't think that gets us anywhere anyway. All of the messing about with limits stuff is a tad beyond my scope anyway.

I take it all back, I've made your proof work!

We have:

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x \, \mathrm{d}x \\ = n\int_0^1 2x \, \mathrm{d}x + (n-1)\int_1^2 2x \, \mathrm{d}x + \cdots + \int_{n-1}^n 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r \int_{n-r}^{n-r+1} 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r(2n-2r+1) \\ = (2n+1)\sum_{r=1}^n r - 2 \sum_{r=1}^n r^2 \\ \Rightarrow 3 \sum_{r=1}^n r^2 = (2n+1)\sum_{r=1}^n r = \frac{1}{2}n(n+1)(2n+1) \\ \Rightarrow \sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1).$

Voila!

We have:

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x \, \mathrm{d}x \\ = n\int_0^1 2x \, \mathrm{d}x + (n-1)\int_1^2 2x \, \mathrm{d}x + \cdots + \int_{n-1}^n 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r \int_{n-r}^{n-r+1} 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r(2n-2r+1) \\ = (2n+1)\sum_{r=1}^n r - 2 \sum_{r=1}^n r^2 \\ \Rightarrow 3 \sum_{r=1}^n r^2 = (2n+1)\sum_{r=1}^n r = \frac{1}{2}n(n+1)(2n+1) \\ \Rightarrow \sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1).$

Voila!

Hurrah! Vielen Danke!

FWoodhouse

I take it all back, I've made your proof work!

We have:

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x \, \mathrm{d}x \\ = n\int_0^1 2x \, \mathrm{d}x + (n-1)\int_1^2 2x \, \mathrm{d}x + \cdots + \int_{n-1}^n 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r \int_{n-r}^{n-r+1} 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r(2n-2r+1) \\ = (2n+1)\sum_{r=1}^n r - 2 \sum_{r=1}^n r^2 \\ \Rightarrow 3 \sum_{r=1}^n r^2 = (2n+1)\sum_{r=1}^n r = \frac{1}{2}n(n+1)(2n+1) \\ \Rightarrow \sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1).$

Voila!

We have:

$\displaystyle \Huge \sum_{r=1}^n r^2 = \sum_{r=1}^n \int_0^r 2x \, \mathrm{d}x \\ = n\int_0^1 2x \, \mathrm{d}x + (n-1)\int_1^2 2x \, \mathrm{d}x + \cdots + \int_{n-1}^n 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r \int_{n-r}^{n-r+1} 2x \, \mathrm{d}x \\ = \sum_{r=1}^n r(2n-2r+1) \\ = (2n+1)\sum_{r=1}^n r - 2 \sum_{r=1}^n r^2 \\ \Rightarrow 3 \sum_{r=1}^n r^2 = (2n+1)\sum_{r=1}^n r = \frac{1}{2}n(n+1)(2n+1) \\ \Rightarrow \sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1).$

Voila!

Shhhhhhhhhhhh. Get to bed!

Hmmm, we might be able to use calculus with a little less work, if we assume the sum of a G.P. Consider:

$\sum_{r=0}^n a^r = \frac{1-a^{n+1}}{1-a}$

Termwise differentiation is okely-dokely, since we're dealing with finite sums (this is important apparently). Differentiating with respect to a, and taking the limit as a→1 then gives:

$\sum_{r=1}^n r = \frac{1}{2}n(n+1)$

Differentiating more times and multiplying by corresponding powers of a gives the natural extensions. Can't take credit though - my brother gave me the idea!

$\sum_{r=0}^n a^r = \frac{1-a^{n+1}}{1-a}$

Termwise differentiation is okely-dokely, since we're dealing with finite sums (this is important apparently). Differentiating with respect to a, and taking the limit as a→1 then gives:

$\sum_{r=1}^n r = \frac{1}{2}n(n+1)$

Differentiating more times and multiplying by corresponding powers of a gives the natural extensions. Can't take credit though - my brother gave me the idea!

Quagmire

Hmmm, we might be able to use calculus with a little less work, if we assume the sum of a G.P. Consider:

$\sum_{r=0}^n a^r = \frac{1-a^{n+1}}{1-a}$

Termwise differentiation is okely-dokely, since we're dealing with finite sums (this is important apparently). Differentiating with respect to a, and taking the limit as a→1 then gives:

$\sum_{r=1}^n r = \frac{1}{2}n(n+1)$

Differentiating more times and multiplying by corresponding powers of a gives the natural extensions. Can't take credit though - my brother gave me the idea!

$\sum_{r=0}^n a^r = \frac{1-a^{n+1}}{1-a}$

Termwise differentiation is okely-dokely, since we're dealing with finite sums (this is important apparently). Differentiating with respect to a, and taking the limit as a→1 then gives:

$\sum_{r=1}^n r = \frac{1}{2}n(n+1)$

Differentiating more times and multiplying by corresponding powers of a gives the natural extensions. Can't take credit though - my brother gave me the idea!

I like that. Very clever.

Hmmm, I don't really understand. Could someone elaborate a bit please?

e-unit

Hmmm, I don't really understand. Could someone elaborate a bit please?

Well, having commended it earlier, I'm not actually all that much of a fan of it now... I hadn't gone through the maths, but now I have, finding that limit takes a fair bit of time, unless there's an easy way I've missed.

$\Large 1 + a + \cdots + a^n = \frac{1 - a^{n+1}}{1-a}$.

Now, differentiating both sides with respect to a,

$\Large 1 + 2a + 3a^2 + ... + na^{n-1} = \frac{(1-a)(-(n+1)a^n) + 1 - a^{n+1}}{(1-a)^2}$.

If you set a=1 on the LHS we get the sum we want, so we need to find the limit of the RHS as a goes to 1.

Let u = 1-a. Then the limit as a goes to 1 is the limit as u goes to 0. Then the RHS is

$\Large \frac{a^n (-n-1 + an) + 1}{(1-a)^2} = \frac{(1+u)^n(un - 1) + 1}{u^2}$

$\Large = \frac{1}{u^2} ( (un-1)( ^nC_0 + ^nC_1 u + \cdots + ^nC_n u^n ) + 1) )$

$\Large = \frac{1}{u^2} ( (un-1)( ^nC_1 u + ^nC_2 u^2 + \cdots + ^nC_n u^n ) + un) )$

$\Large = \frac{1}{u^2} ( (un-1)( ^nC_2 u^2 + ^nC_3 u^3 + \cdots + ^nC_n u^n ) + u^2 n^2) )$

$\Large = (un-1)( ^nC_2 + ^nC_3 u + \cdots + ^nC_n u^{n-2} ) + n^2)$

Then the limit of this as u goes to 0 is

$\Large n^2 - ^nC_2 =n^2 -\frac{1}{2}n(n-1) = \frac{1}{2}n(n+1)$

I really hope I've missed a trick which makes it much easier...

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