series!!

[homad2000]

1. The problem statement, all variables and given/known data

consider the series: a + (a+d) + (a+3d) + (a+6d) + (a+10d) + (a+15d) .....
find a formula for nth term, and the sum of the first n terms.

2. Relevant equations

I think, it is similar to the Fibonacci series.


3. The attempt at a solution

well, I tried rearange and simplify it, but no clue!!

[tiny-tim]

hi homad2000! :wink:

well, first, what is the formula for nth term? :smile:

[SteamKing]

Hint: Look at how each term in the series differs from the last. From the partial series given, you can assume that the quantity 'a' is in each term of the series. Now look at how the part with the quantity 'd' changes depending on which term is considered. In the first term, there is no 'd'. In the second term, a single 'd'. In the third term, '3d'. The coefficient of the 'd' term is some function of the 'i'th term of the series.

[homad2000]

OK, I see how this series working, the nth term can be found like this:

a(n) = a(n-1) + (n-1)d

but how about the sum of the series?

[tiny-tim]

a(n) = a(n-1) + (n-1)d

nooo … try again :smile:

[homad2000]

nooo … try again :smile:

why? if we want to get for example the 4th term, it's the third term + (4-1)d = (a+3d) + 3d = a+6d ?

[tiny-tim]

oh sorry, i misread your a(n-1) as a product :redface:

ok now what is an in absolute terms, not as a function of an-1 ? :smile:

[homad2000]

great!

I got: a(n) = a + (n^2 - n ) / 2 * d !!

any hints how to start solving the second part?

[tiny-tim]

hi homad2000! :smile:

(try using the X2 icon just above the Reply box :wink:)

ok :smile:

now sum each bit separately …

∑ a is easy! :tongue2: …

for ∑ (n2 - n)/2, rewrite that as ∑ n(n-1)/2 …

does that remind you of anything? :wink:

[homad2000]

:) hahah, i wasnt thinking that way!! anyways, thank you very much!!