Calculating the Sum of n(n+1) Using Even Numbers and Odd Number Sums

[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.

 

[benorin]

Here is a link to the answer of your question I had posted on another thread:

https://www.physicsforums.com/showthread.php?t=97842

 

[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}

 

[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?

 

[benorin]

The proof ? Oh, just I used Maple .

 

[MathematicalPhysicist]

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

 

[George Jones]

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.)

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

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,

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

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

Regards,
George

 

[HallsofIvy]

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

 

[George Jones]

I would think that it would be simpler to write
\begin{equation*}\begin{split}\sum_{k=1}^{n} k \left( k + 1 \right) \\&amp;= \sum_{k=1}^{n}(k^2+ k)\\&amp;= \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]

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.