Find the sum.

n

E(2-5i) =

i=1

how would i do this?

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- Nov 7th 2009, 04:49 PMSneakysummation problem
Find the sum.

n

**E**(2-5i) =

i=1

how would i do this? - Nov 7th 2009, 05:02 PMDebsta
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 PMSneaky
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 PMJG89
$\displaystyle \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 PMDebsta
Formula for sum of AP is (n/2)(2a+(n-1)d)

- Nov 7th 2009, 05:46 PMSneaky
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 PMDebsta
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 PMSneaky
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 PMDebsta
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 PMSneaky
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 PMDebsta
(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 PMSneaky
it helped and i got it right , i got 5^100 as the answer

- Nov 7th 2009, 06:32 PMVonNemo19
- Nov 7th 2009, 06:35 PMDebsta
I think the answer should be 5^100 minus 1.

- Nov 7th 2009, 06:38 PMVonNemo19