## Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

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### Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

Please may s.o. to help me to solve the sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000.
japiga

Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

japiga wrote:Please may s.o. to help me to solve the sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000.

These products are what is called "rectangular" numbers because, if you draw the products in "dot" form, you get rectangles. For instance:

Code: Select all
2:  **6:  ***    ***12: ****    ****    ****

What have you found when you've tried to find a pattern? For instance:

2 = 2
2 + 6 = 8
2 + 6 + 12 = 20
2 + 6 + 12 + 20 = 40
2 + 6 + 12 + 20 + 30 = 70

maggiemagnet

Posts: 298
Joined: Mon Dec 08, 2008 12:32 am

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

geometric progression?
japiga

Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

But how to put it in mathematic formula? I have no idie at this moment.
japiga

Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

japiga wrote:geometric progression?

To be a geometric series, there would have to be a common ratio. Is there?

What did you find at the link in my previous post?

maggiemagnet

Posts: 298
Joined: Mon Dec 08, 2008 12:32 am

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

use...

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

and

$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{999}i(i+1)=\sum_{i=1}^{999}(i^2+i)=\sum_{i=1}^{999}i^2+\sum_{i=1}^{999}i=\cdots$

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

Dear Martingale. T hose sums are not sums of orders: geometric and arithmetic sum of order. Please give me more details it is very important to me. How to calculate sum of i^2 = 1^2+2^2+3^2…..+ 999^2 and sum of i = 1 + 2 +3 + …. + 999. Hot to calculate it? I have to know exact sum number. It is huge number of ciphers!!? Please help me!
japiga

Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

japiga wrote:Dear Martingale. T hose sums are not sums of orders: geometric and arithmetic sum of order. Please give me more details it is very important to me. How to calculate sum of i^2 = 1^2+2^2+3^2…..+ 999^2 and sum of i = 1 + 2 +3 + …. + 999. Hot to calculate it? I have to know exact sum number. It is huge number of ciphers!!? Please help me!

I gave you the formulas for the sum of $i$ and $i^2$

Just plug 999 into the formulas

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

the first one is clear, it is arithmetic order: 1 + 2 + 3 + ... + 999, since d = 1 and it is easy to calculate, but what is with another one 1^2+2^2+3^2…..+ 999^2. It is not geometric progression order. How to calculate this one?
japiga

Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

### Re: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000

japiga wrote:the first one is clear, it is arithmetic order: 1 + 2 + 3 + ... + 999, since d = 1 and it is easy to calculate, but what is with another one 1^2+2^2+3^2…..+ 999^2. It is not geometric progression order. How to calculate this one?

Like I have above...

$\sum_{i=1}^{n}i^2=1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$

your $n=999$

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

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