Summation Notation problem

[rsgunter21]

I'm having issues with this problem, being that there are no examples in the book that come close to it. I will do the best I can to make it look like something...


Find:

2500
E j
j=1

E = Sigma.

Thanks,
Ryan

 

[pka]

\(\displaystyle \L
\sum\limits_{j = 1}^N j = \frac{{N\left( {N + 1} \right)}}{2}\)

 

[rsgunter21]

\(\displaystyle \L
\sum\limits_{j = 1}^N j = \frac{{N\left( {N + 1} \right)}}{2}\)

awesome, but why exactly does it work like that?

thanks

 

[daon]

\(\displaystyle \L
\sum\limits_{j = 1}^N j = \frac{{N\left( {N + 1} \right)}}{2}\)

awesome, but why exactly does it work like that?

thanks

The summation is equal to: 2500 + 2499 + 2498 + ... + 2 + 1

If we place this summation aside itself in reverse order and add each row together you'll see:

2500 + 1 = 2501
2499 + 2 = 2501
2498 + 3 = 2501
.
.
.
2 + 2499 = 2501
1 + 2500 = 2501

And we have 2500 2501's, so Twice the sum is equal to 2500(2501). Divide both sides by two and you get the Sum is equal to half of (2500)(2501).

Or (2500)(2500 + 1)/2

-Daon