1. Good post? |

Need Fomulas

Hi All,

Can some one please give me the formula to

say 5+6+7+.........+1000 = ?
5*6*7...........*1000 = ?
say 5+10+15+20.........+1000 = ?
5*10*20*40*80*160*320 = ?

is there any formula to find product of this series
5*10*15*20*......*1000

Thanks a lot for your time

2. Good post? |
5*10*15*20*......*1000
= 5*1*5*2*5*3........5*200
= 5^200 * (1*2*3....200)
= 5^200*200!

Normally I dont think they will ask you to calculate the actual value. The asnwer will be a variant of a factorial and a power. The reason is Official Guide doesnt contain any explanations or formulae, so they wont directly test you on calculations.

n*(n+1)/2 = sum of n numbers. So, if you say 5+6+7....1000 it means you will need to calculate for 1+2+3....1000 then subtract 1+2+3+4+5 from it.

Any other opinions?

3. Good post? |
repear.

4. Good post? |
adding my 2cents, formula to find sum of first n squares = 1 + 22 + 32 + ... + n2 =

5. Good post? |
Originally Posted by mxr55820
Hi All,

Can some one please give me the formula to

say 5+6+7+.........+1000 = ?
5*6*7...........*1000 = ?
say 5+10+15+20.........+1000 = ?
5*10*20*40*80*160*320 = ?

is there any formula to find product of this series
5*10*15*20*......*1000

Thanks a lot for your time
I typically try to dissuade student from memorizing a lot of formulas. For one thing, they seldom add much value on their own.

Now, for the GMAT, I do happen to think there is some value in memorizing the sum of the first n positive integers.

That is: 1+2+3+4+5+....+n = n(n+1)/2
For example, the sum of the first 20 positive integers (1+2+3+...+20) is equal to 20(21)/2, which equals 210

Having said that, it is unlikely that you will be given a GMAT question that asks "what is the sum of integers from 1 to 20 inclusive?"

However, you may see something like, "What is the sum of all positive integers less than 701 that are divisible by 7?"

In other words, we want to find: 7+14+21+28+...+700

Well, if we factor out a 7, we get 7+14+21+28+...+700= 7(1+2+3+4+....100)

At this point, we can use the formula to see that (1+2+3+4+....100) = 5050

So, 7(1+2+3+4+....100) = 7(5050) = 35350

Cheers,
Brent

6. Good post? |

Another formula

Also regardless of deviation of the arithmetic series:

Sum=n(a1+an)/2

works for 1,2,3... and 7,14,21 ...

SPOILER: mouseover variables

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•

SEO by vBSEO ©2010, Crawlability, Inc.