Finding the sum of

1. Jan 15, 2006

MathematicalPhysicist

finding the sum of....

1*2+2*3+3*4+...+n(n+1)

i need to find to what it's equal?

i tried doing various ways:

2*(1+3)+4*(3+5)+6(5+7)+...

which means an even number times the sum of two odd numbers
so let's say:
a=2m
b=2m-1

a(b+3)+a(b+5)+a(b+7)...+n(n+1)
ab*n+a(3+5+7...+2n-1)*(n-1)=
a{[bn+(n-1)(3+5+7+...2n-1)]}

when 3+5+7+...(2n-1)=(3+s_n)n/2
when s_n=3+n(2n-1-3)=2n^2-4n+3
so (3+s_n)n/2=(2n^2-4n+6)n/2=n^3-2n^2+3n

here i pretty much given up, can someone help me.

2. Jan 15, 2006

benorin

3. Jan 15, 2006

benorin

In particular,

$$\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}$$

4. Jan 15, 2006

MathematicalPhysicist

i think you jumped there too quickly, here:
sigma[k=1 to n](k+j)!/(k-1)!=n(n+j+1)!/(j+2)n!

the sigma equals:
(1+j)!/0!+(2+j)/1!!+...+(n+j)!/(n-1)!

so where's the thing in between?

5. Jan 15, 2006

benorin

The proof ? Oh, just I used Maple .

Last edited: Jan 15, 2006
6. Jan 15, 2006

MathematicalPhysicist

so does someone know how to prove it, or should i use my bare hands. (-:

7. Jan 15, 2006

George Jones

Staff Emeritus
Now that you know the final answer, the result should be easy to prove by mathematical induction.

Also...
This method works! (For an even number of terms.)

$$\begin{equation*} \begin{split} 1*2 + 2*3 + 3*4 + 4*5 + 5*6 + 6*7 + 7*8 + 8*9 + \cdots &= 2 \left( 1 + 3 \right) + 4 \left( 3 + 5 \right) + 6 \left( 5 + 7 \right) + 8 \left( 7 + 9 \right) + \cdots\\ &= 8 + 32 + 72 + 128 + \cdots\\ &= 8 \left( 1 + 4 + 9 + 16 + \cdots \right)\\ &= 8 \left( 1 + 2^{2} + 3^{2} + 4^{2} + \cdots \right)\\ \end{split} \end{equation*}$$

The stuff in the brackets is a known series, but this is not a proof. It does, however, suggest the following more formal method. Let $n = 2m$, so that

$$\sum_{k=1}^{2m} k \left( k + 1 \right) = \sum_{k=1}^{2m} a_{k},$$

where $a_{k} = k \left( k + 1 \right)$. Define a series by $b_{1} = a_{1} + a_{2}$, $b_{2} = a_{3} + a_{4}$, etc. In other words, $b_{j} = a_{2j - 1} + a_{2j}$.

Now,

$$\begin{equation*} \begin{split} \sum_{k=1}^{2m} k \left( k + 1 \right) &= \sum_{k=1}^{2m} a_{k}\\ &= \sum_{j=1}^{m} b_{j}\\ &= \sum_{j=1}^{m} \left[ a_{2j-1} + a_{2j} \right] .\\ \end{split} \end{equation*}$$

What do you get when you sub in for the a's?

Regards,
George

Last edited: Jan 15, 2006
8. Jan 15, 2006

HallsofIvy

Staff Emeritus
I would think that it would be simpler to write
$$\begin{equation*}\begin{split}\sum_{k=1}^{n} k \left( k + 1 \right) \\&= \sum_{k=1}^{n}(k^2+ k)\\&= \sum_{k=1}^{n} k^2+ \sum_{k=1}^{n} k\end{split}\end{equation*}$$
And those two sums are well known.

9. Jan 15, 2006

George Jones

Staff Emeritus
Yikes!

This exemplifies something I often tell people: take care when replacing symbols by numbers, because this sometimes complicates the issue at hand.

Mea culpa.

Regards,
George

10. Jan 16, 2006

MathematicalPhysicist

thank, i believe this is much easier from what i had thought before.

the first k^2 sum is of what galileo find, and the second one is just plane arithmatic progression.

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