# summation problem

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• Nov 7th 2009, 04:49 PM
Sneaky
summation problem
Find the sum.
n
E (2-5i) =
i=1

how would i do this?
• Nov 7th 2009, 05:02 PM
Debsta
Generate the first few terms to see what's going on.
Put in i=1, i=2, i=3,...
The terms are -3, -8, -13, -18 etc. and you want to add these up.
The numbers form an AP (arithmetic prgression) so find a and d, and use the sum of an AP formula.
• Nov 7th 2009, 05:06 PM
Sneaky
i think i get what your saying and i get n(-3-5)
but its still wrong
can u show me the steps and the solution to solve this
• Nov 7th 2009, 05:12 PM
JG89
$\sum_{i = 1}^n 2 - 5i = \sum_{i = 1}^n 2 - 5 \sum_{i = 1}^n i = 2 - 5 \sum_{i = 1}^n i$.

Now use the formula for the sum of the first n positive integers.
• Nov 7th 2009, 05:13 PM
Debsta
Formula for sum of AP is (n/2)(2a+(n-1)d)
• Nov 7th 2009, 05:46 PM
Sneaky
i am still very confused on how to do this question
what is the a and d
and how did u get (n/2)(2a+(n-1)d)
• Nov 7th 2009, 05:50 PM
Debsta
Have you learnt about arithmetic progressions? If not check out JG89's suggestion. If you have learnt about APs the you should know the sum of a AP formula.
• Nov 7th 2009, 05:57 PM
Sneaky
ok now i get this question but now i have this question

n
E (i+1)(i+2)
i=1

the GP formula is

a(1-r^n)
---------
1-r

what would a and r be
a is the starting point so it would be 6 ?
• Nov 7th 2009, 06:18 PM
Debsta
List a few terms first to get a feel for what you are looking at. Yes the first term is 6. list a few more. Is it a GP??
• Nov 7th 2009, 06:24 PM
Sneaky
ok i got that question now i dont know how to do this one

100
E (5^(i) - 5^(i-1))
i=1
• Nov 7th 2009, 06:28 PM
Debsta
(5^(i) - 5^(i-1)) = 5^(i) - 5^(i) x 5^(-1) = 5^(i)[ 1 - 5^(-1)]
That should help.
• Nov 7th 2009, 06:31 PM
Sneaky
it helped and i got it right , i got 5^100 as the answer
• Nov 7th 2009, 06:32 PM
VonNemo19
Quote:

Originally Posted by Sneaky
ok i got that question now i dont know how to do this one

100
E (5^(i) - 5^(i-1))
i=1

Do you know what a telescoping sum is?

It means that all of the middle terms cancel leaving only a couple at the beggining and at the end. Think about your problem, which terms are gonna cancel other terms, and what are the only terms that will remain standing?
• Nov 7th 2009, 06:35 PM
Debsta
I think the answer should be 5^100 minus 1.
• Nov 7th 2009, 06:38 PM
VonNemo19
Quote:

Originally Posted by Debsta
I think the answer should be 5^100 minus 1.

And that would be correct!

(Wink)
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