# Finding the sum of

#### MathematicalPhysicist

Gold Member
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.

Homework Helper

#### benorin

Homework Helper
In particular,

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

#### MathematicalPhysicist

Gold Member
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?

#### benorin

Homework Helper
The proof ? Oh, just I used Maple .

Last edited:

#### MathematicalPhysicist

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

#### George Jones

Staff Emeritus
Gold Member
loop quantum gravity said:
so does someone know how to prove it, or should i use my bare hands. (-:
Now that you know the final answer, the result should be easy to prove by mathematical induction.

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

#### HallsofIvy

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.

#### George Jones

Staff Emeritus
Gold Member
HallsofIvy said:
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.
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

#### MathematicalPhysicist

Gold Member
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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