Complicated algebra in order to prove a sequence

**Jeuk** · February 2nd 2009, 02:58 AM

Okay, so since I don't know how to use the math wraps for signs, etc, I'll just try to explain this.

We just got into sequences, and after the professor showed us the proof to find a formula for the sum of a sequence where i (infront of the sigma) is squared, he challanged us to try and resolve the the proof for the cubed sequence.

Just to explain what this looks like since I don't know the signage:

A sigma with:
i=1 on the bottom (where it starts, if I haven't already forgotten.)
n on the top (where it ends)
And to its right: i cubed. (i^3)

Then I ended up with a bunch of polynomial functions. I remember understanding it in class, but my poor note taking and quick-lost memory failed me once more, so I am not sure how I got here, but I am pretty sure it's correct. I ended up with this:

f(1) = A + B + C + D + E = 1
f(2) = 16A + 8B + 4C +2D + E = 9
f(3) = 81A + 27B + 9C + 3D + E = 37
f(4) = 256A + 64B + 16C + 4D + E = 37
f(5) = 625A + 125B + 25C + 5D + E = 226

Now, I'm pretty sure, as far as I remember from class, that all we had to do was algebraically solve for all of these using whatever method we could. I've tried a few that I can swing but kept getting stuck.

Perhaps someone can give me a hint as how to go about this, maybe a simpler way. Or maybe I'm just doing it wrong and there is a polynomial calculator that'll do the algebra. Kind of embarrassing and maybe this shouldn't even be on the precalc forum. Either way, I'd like to hear your advice, thanks.

**Soroban** · February 2nd 2009, 09:12 AM

Hello0, Jeuk!

Your work is great . . . Thank you for showing it!

Find a formula for: . $\sum^n_{i=1}i^3$

We crank out the first few partial sums:

. . $\begin{array}{ccc}1^3 &=& 1 \\ 1^3+2^3 &=& 9 \\ 1^3+2^3+3^3 &=&36 \\ 1^3+2^3+3^3+4^3 &=&100 \\ 1^3+2^3+3^3+4^3+5^3 &=&225 \end{array}$

We conjecture that the formula is of the form: . $f(n) \:=\:An^4 + Bn^3 + Cn^2 + Dn + E$

Then I stack the equations (in reverse order):

. . $\begin{array}{ccccccc} f(5) &=& 625A + 125B + 25C + 5D + E &=& 225 & [1] \\ f(4) &=& 256A + 64B + 16C + 4D + 3 &=& 100 & [2] \\ f(3) &=& 81A + 27B + 9C + 3D + E &=& 36 & [3] \\ f(2) &=& 16A + 8B + 4C + 2D + e &=& 9 & [4] \\ f(1) &=& A + B + C + D + E &=& 1 & [5] \end{array}$

There is a nice "rhythm" for solving this system of equations.
I hope you can catch on to the pattern . . .

$\begin{array}{cccccc}\text{Subtract [1]-[2]} & 369A + 61B + 9C + D &=& 125 & [6] \\ \text{Subtract [2]-[3]} & 175A + 37B + 7C + D &=& 64 & [7] \\ \text{Subtract [3]-[4]} & 65A + 19B + 5C + D &=& 27 & [8] \\ \text{Subtract [4]-[5]} & 15A + 7B + 3C + D &=& 8 & [9] \end{array}$

$\begin{array}{ccccc}\text{Subtract [6]-[7]} &194A + 24B + 2C &=& 61 & [10] \\ \text{Subtract [7]-[9]} & 110A + 18B + 2C &=& 37 & [11] \\ \text{Subtract [8]-[9]} & 50A + 12B + 2 &=& 19 & [12] \end{array}$

$\begin{array}{ccccc}\text{Subtract [10]-[11]} & 84A + 6B &=& 24 & [13] \\ \text{Subtract [11]-[12]} & 60A + 6B &=& 18 & [14] \end{array}$

$\text{Subtract [13]-[14]}\quad 24A \:=\:6 \quad\Rightarrow\quad\boxed{ A \:=\:\tfrac{1}{4}}$

Substitute into [14]: . $60\left(\tfrac{1}{4}\right) + 6B \:=\:18 \quad\Rightarrow\quad \boxed{B \:=\:\tfrac{1}{2}}$

Substitute into [12]: . $50\left(\tfrac{1}{4}\right) + 12\left(\tfrac{1}{2}\right) + 2C \:=\:19 \quad\Rightarrow\quad\boxed{ C \:=\:\tfrac{1}{4}}$

Substitute into [9]: . $15\left(\tfrac{1}{4}\right) + 7\left(\tfrac{1}{2}\right) + 3\left(\tfrac{1}{4}\right) + D \:=\:9 \quad\Rightarrow\quad\boxed{D \:=\:0}$

Substitute into [1]: . $\tfrac{1}{4} + \tfrac{1}{2} + \tfrac{1}{4} + 0 + E \:=\:1 \quad\Rightarrow\quad\boxed{E \:=\:0}$

Hence: . $f(n) \;=\;\tfrac{1}{4}n^4 + \tfrac{1}{2}n^3 + \tfrac{1}{4}n^2 \;=\;\frac{n^2}{4}(n^2+2n+1)$

. . Therefore: . $f(n) \;=\;\frac{n^2(n+1)^2}{4}$

**Jeuk** · February 2nd 2009, 01:30 PM

Wow, flawless explanation.
Thank you for sharing your knowledge.

I'll try and keep that nifty trick in mind next time I have many polynomials.

Thanks again.

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