# Math Help - The sum of a most problematic sequence.

1. ## The sum of a most problematic sequence.

I've been trying this question and I keep getting stuck!

The odd thing is that I can't see what's wrong.

I get an expression at the end but it doesn't work when I substitute numbers in. I'm pretty sure that my expression for 1^2+2^2+3^2+......+n^2 is correct, but i'm not certain.

2. Hello !

The idea is good ! (I wonder how you got it)
however, I can point out two problems :

$(i+1)^4-i^4=4i^3+6i^2+4i+1$
In your second line, you forgot the $-i^4$ part, but I guess it is just a typo.

When substituting different values of $i$, you wrote 6(1), 6(2)... which is false since it is 4i in the formula. Moreover, you forgot each time to add 1.

So adding them together shall give something like :

$(n+1)^4-1^4=4(1^3+2^3+\dots+n^3)+6(1^2+2^2+\dots+n^2)+{\co lor{red}4}(1+2+\dots+n)+{\color{red}n(1)}$

Edit : and yes, your expression for the sum of the squares is correct =)

3. Originally Posted by Showcase_22
I've been trying this question and I keep getting stuck!

The odd thing is that I can't see what's wrong.

I get an expression at the end but it doesn't work when I substitute numbers in. I'm pretty sure that my expression for 1^2+2^2+3^2+......+n^2 is correct, but i'm not certain.

There is an another (easier?) method. You need to know that the sum of the first n k-th powers is a polynomial of degree k+1 in n (because the (k+1)th differences of the sequence of sums are all zero, or equivalently the k-th differences are constant).

So in this case you are looking for a quartic in n. A general quartic has five coefficients, so if you eveluate the sum for n=1, 2, 3, 5, 5 you will have five linear equations in the coefficients to solve.

RonL

4. Originally Posted by CaptainBlack
There is an another (easier?) method. You need to know that the sum of the first n k-th powers is a polynomial of degree k+1 in n (because the (k+1)th differences of the sequence of sums are all zero, or equivalently the k-th differences are constant).

So in this case you are looking for a quartic in n. A general quartic has five coefficients, so if you eveluate the sum for n=1, 2, 3, 5, 5 you will have five linear equations in the coefficients to solve.

RonL
ummm....

Could you give me an example??

I didn't come up with this method on my own, it's a question that has various stages. This was the last part!

I'll take on board what you wrote Moo. It seems that I just wasn't paying attention or something (I think my sister was watching Resident Evil on maximum volume!).

5. Originally Posted by CaptainBlack
There is an another (easier?) method. You need to know that the sum of the first n k-th powers is a polynomial of degree k+1 in n (because the (k+1)th differences of the sequence of sums are all zero, or equivalently the k-th differences are constant).

So in this case you are looking for a quartic in n. A general quartic has five coefficients, so if you eveluate the sum for n=1, 2, 3, 5, 5 you will have five linear equations in the coefficients to solve.

RonL
We know that:

$S(n)=\sum_{k=1}^n k^3=a+bn+cn^2+dn^3+en^4$

for some $a,\ b,\ c,\ d,\ e$

Now:

$S(1)=1=a+b+c+d+e$

$S(2)=1+2^3=9=a+2b+4c+8d+16e$

$S(3)=S(2)+3^3=36=a+3b+9c+27d+81e$

$S(4)=S(3)+4^3=100=a+4b+16c+64d+256e$

$S(5)=S(4)+5^3=225=a+5b+25c+125d+625e$

Which has solution $a=b=0,\ c=1/4,\ d=1/2,\ e=1/4$

So:

$S(n)=\frac{n^2(n+1)^2}{4}$

RonL

6. ## Difference equations

I will show a new way for the solution.
This may take a little bit more time but may be this will help you in some other subjects.
We shall build a difference equation by the given relation, then we will attempt to solve it.

we are asked to find what is $1^{3}+\cdots+n^{3}$?
Set $u(n):=1^{3}+\cdots+n^{3}$, then we see that $u(n+1)=(n+1)^{3}+u(n)$ holds and the initial condition is $u(1)=1$.
By using the forwards difference operator $\Delta$, we can write the following difference equation
$\Delta u(n)=(n+1)^{3}\text{ with }u(1)=1,\qquad(1)$
where $\Delta u(n):=u(n+1)-u(n).$

Solving (1) wil give the answer.
But how to solve it is the main question.
For this purpose, we need the definition of the factorial function as follows:
$n^{(k)}:=n(n-1)\ldots(n-k+1).$
This notation helps us very much because
$\Delta n^{(k)}=kn^{(k-1)}\text{ for }k\in\mathbb{N}$, which can be regarded as discrete corresponding to derivative of the polynomials.
It can be easily shown that the followings are true:
$n=n^{(1)}$
$n^{2}=n^{(2)}+n^{(1)}$
$n^{3}=n^{(3)}+3n^{(2)}+n^{(1)}$

Then, we can rewrite (1) in the following form:
$\Delta u(n)=n^{3}+3n^{2}+3n^{1}+1=n^{(3)}+6n^{(2)}+7n^{(1 )}+1.\qquad(2)$
Summing up (2) from $1$ to $n-1$, we find that
$u(n)-u(1)=\sum\limits_{i=1}^{n-1}\Delta u(i)=\sum\limits_{i=1}^{n-1}[i^{(3)}+6i^{(2)}+7i^{(1)}+1]$
.................................._ $=\bigg[\frac{i^{(4)}}{4}+6\frac{i^{3}}{3}+7\frac{i^{(2)}} {2}+i^{(1)}\bigg]_{i={\color{red}{1}}}^{{\color{blue}{n-1}}}$
.................................._ $={\color{blue}{\frac{1}{4}(n-1)^{(4)}+2(n-1)^{(3)}+\frac{7}{2}(n-1)^{(2)}+(n-1)^{(1)}}}-{\color{red}{1}}.$
Simplifying the last line by the notation of the factorial function and considering the initial function, we get
$u(n)=\frac{1}{4}n^{4}-\frac{1}{2}n^{3}-\frac{1}{4}n^{2}$
......_ $=\frac{1}{4}n^{2}(n+1)^{2},$
which should be the desired solution for (1).