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

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In summary, the conversation discussed different approaches to finding the sum of 1*2+2*3+3*4+...+n(n+1) and ended with a simpler method of breaking it down into two well-known sums, the sum of squares and the sum of an arithmetic progression.
  • #1
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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.
 
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  • #3
In particular,

[tex]\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}[/tex]
 
  • #4
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
The proof ? Oh, just I used Maple :wink: .
 
Last edited:
  • #6
so does someone know how to prove it, or should i use my bare hands. (-:
 
  • #7
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.)

[tex]
\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*}
[/tex]

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 [itex]n = 2m[/itex], so that

[tex]\sum_{k=1}^{2m} k \left( k + 1 \right) = \sum_{k=1}^{2m} a_{k},[/tex]

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

Now,

[tex]
\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*}
[/tex]

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

Regards,
George
 
Last edited:
  • #8
I would think that it would be simpler to write
[tex]\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*}[/tex]
And those two sums are well known.
 
  • #9
HallsofIvy said:
I would think that it would be simpler to write
[tex]\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*}[/tex]
And those two sums are well known.

Yikes! :blushing:

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
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.
 

1. What is the formula for calculating the sum of n(n+1) using even numbers and odd number sums?

The formula for calculating the sum of n(n+1) using even numbers and odd number sums is (n/2)(n+1) + (n+1)^2/2. This formula applies for any value of n, where n is a positive integer.

2. How do even numbers and odd number sums play a role in this calculation?

Even numbers and odd number sums are used in this calculation because they are the building blocks of the formula. The sum of n consecutive even numbers is equal to n(n+1), and the sum of n consecutive odd numbers is equal to (n+1)^2. Therefore, by combining these two sums, we can arrive at the formula for calculating the sum of n(n+1) using even numbers and odd number sums.

3. Can this formula be used for any value of n?

Yes, this formula can be used for any value of n as long as n is a positive integer. It is a general formula that applies to all positive integers, and it will always give the correct result.

4. Is there a simpler way to calculate the sum of n(n+1) using even numbers and odd number sums?

Yes, there is a simpler way to calculate the sum of n(n+1) using even numbers and odd number sums. Instead of using the formula, you can use the shortcut method of multiplying n by (n+1) and then dividing the result by 2. This method will give you the same result as the formula and is easier to calculate mentally.

5. How can this formula be applied in real-life situations?

This formula can be applied in various real-life situations, such as calculating the number of pairs of shoes needed for a group of people, finding the total number of books in a library, or determining the number of students in a school based on the number of classes. It can also be used in mathematical and scientific calculations that involve consecutive numbers.

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