Sum of a series

  • #1
1,444
2

Homework Statement


Find a general formula for
[tex] (1\cdot 2) +(2\cdot 3) +... + (n\cdot (n+1)) [/tex]

Homework Equations


The sum of the first n integers
[tex] 1+2+3+... + n = \frac{n(n+1)}{2} [/tex]

The Attempt at a Solution


i thought of doing this
[tex] (1+2+3...) * (2+3+4+...) [/tex]

but then the problem of cross terms comes about

additionally this is not an arithmetic series nor a geometric series because the common difference increases by 1 each term and there is no common factor.

how wouldi get rido of the cross terms??
do i even need to do it this way?? Another method is possible??

Please help!!

thanks!
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
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Do you know the formula for [tex]\sum_{i=1}^n i^2[/tex]?
 
  • #3
caffeine
So you want to sum:

1*2 + 2*3 + 3*4 + 4*5 + ... + n(n+1)

It's very difficult working with numbers. You'll never get anywhere. So the first order of business is to rewrite the above sum using a symbolic sum:

[tex]
1*2 + 2*3 + 3*4 + 4*5 + ... + n(n+1) = \sum_{i=start}^{end} foo
[/tex]

You shouldn't have too much difficulty figuring out what "start", "end" and "foo" should be to represent the left side of that equation.

You'll see that "foo" is a product of two terms. They'll look "factored". Unfactor them (distribute terms) and you'll end up with a sum of two terms. It'll look like:

[tex]
\sum_{i=start}^{end} hello + kitty
[/tex]

One term is Quasar's term. The other term you'll want to ask Mr. Gauss about.

Note: You can sum some of the series some of the time, but you can't sum all of the series all of the time. This case happens to be one of the "some of the time".
 
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  • #4
1,444
2
so then the sum of the series is

[tex] \sum_{i=1}^{n} i(i+1) = \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i [/tex]

yaay

thanks guys
 

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