1. summation problem

Find the sum.
n
E (2-5i) =
i=1

how would i do this?

2. 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.

3. 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

4. $\displaystyle \sum_{i = 1}^n 2 - 5i = \sum_{i = 1}^n 2 - 5 \sum_{i = 1}^n i = 2 - 5 \sum_{i = 1}^n i$.

Now use the formula for the sum of the first n positive integers.

5. Formula for sum of AP is (n/2)(2a+(n-1)d)

6. 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)

7. 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.

8. 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 ?

9. 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??

10. ok i got that question now i dont know how to do this one

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

11. (5^(i) - 5^(i-1)) = 5^(i) - 5^(i) x 5^(-1) = 5^(i)[ 1 - 5^(-1)]
That should help.

12. it helped and i got it right , i got 5^100 as the answer

13. 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?

14. I think the answer should be 5^100 minus 1.

15. Originally Posted by Debsta
I think the answer should be 5^100 minus 1.
And that would be correct!

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