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

Please may s.o. to help me to solve the sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000.
by japiga
on Mon Sep 20, 2010 3:42 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

geometric progression?
by japiga
on Tue Sep 21, 2010 8:03 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

But how to put it in mathematic formula? I have no idie at this moment.
by japiga
on Tue Sep 21, 2010 8:04 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

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...
by japiga
on Wed Sep 22, 2010 1:22 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

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?
by japiga
on Wed Sep 22, 2010 1:49 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

1^2+2^2+3^2…..+ 999^2 . How to put this sum in the simple mathematic formula? Since, only 999^2 = 998001and imagine total sum: + 998^2 + 997^2 + 996^2 + ... + 1. I hope that now is more clear what I actually need. Thanks a lot!
by japiga
on Wed Sep 22, 2010 2:10 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

Do you think that this assignment is finished if I accept this and moreover if my teacher accept it? Is there more mathematic approach and more predictable the final result of this “quadratic” sum? Since first sum (1+2+3+…+999) is more clear how to reach the final result!
by japiga
on Wed Sep 22, 2010 2:26 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

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

Now, it's OK. I understood. It sounds very simple. Thanks a lot. This is great way of communication and very quick tool for maths problem solving! Be in touch. Regards
by japiga
on Thu Sep 23, 2010 8:03 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Sum of order: 1*2 + 2*3+ 3*4+....+ 999*1000
Replies: 14
Views: 5958

7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)  TOPIC_SOLVED

Find the least natural number n, so that anyone bellow divisions may not be shorten:

7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

I found that such number n = 17, but how to prove it?
by japiga
on Thu Sep 23, 2010 8:28 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)
Replies: 15
Views: 4387

Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)  TOPIC_SOLVED

Yes, sorry. "shorten" means "divided not to be further divided". Or generally, not to be divided at all. Is it now more clear?
by japiga
on Thu Sep 23, 2010 11:44 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)
Replies: 15
Views: 4387
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