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Arithmetic progressions

The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


I don't have a clue! All I've got is sn=n(n+2)
Reply 1
Avatar for kkboyk
kkboyk
21
Original post by Lucofthewoods
The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


I don't have a clue! All I've got is sn=n(n+2)


Use the formula for the term of the first n terms of a series, and you're given to find the first three terms so when n=3. Put it in your formula to get the answer.
Reply 2
Avatar for joostan
joostan
13
Original post by Lucofthewoods
The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


I don't have a clue! All I've got is sn=n(n+2)

Let the nth term of your series be given by unu_n and the sum of the first nn terms be SnS_n.
Note that: S1=u1S_1=u_1.

You know a formula for SnS_n. So if you sub in n=1n=1 you can find u1u_1.

What can you say about S2S_2 and its relation to S1S_1 and u2u_2?
What about S3S_3?
(edited 7 years ago)
Reply 3
Avatar for Naruke
Naruke
11
Original post by Lucofthewoods
The sum of the first n terms of a series is n(n+2). Find the first three terms of the series.


I don't have a clue! All I've got is sn=n(n+2)


Call your arithmetic progression an a_n . a1=a a_1 = a , a2=a+d a_2 = a + d , a3=a+2d a_3 = a+2d , an=a+(n−1)d a_n = a + (n-1)d . The partial sums of this progression, aka the sum of the first n terms, is found by adding up the first n terms of our arithmetic progression. a1+a2+a3+a4+...+an a_1 + a_2 + a_3 + a_4 + ... + a_n this can be compacted by using sigma notation ∑n=1nan=a1+a2+a3+a4+...+an \sum\limits_{n=1}^n a_n = a_1 + a_2 + a_3 + a_4 + ... + a_n . We are told what the sum of the first n n terms of our progression is n(n+2) n(n+2) . So, Sn=∑n=1nan=n(n+2) S_n= \sum\limits_{n=1}^n a_n = n(n+2)

Notice, that when n=1,2,3 n = 1,2,3

S1=∑n=11an=a1 S_1 = \sum\limits_{n=1}^1 a_n = a_1
S2=∑n=12an=a1+a2 S_2 = \sum\limits_{n=1}^2 a_n = a_1 + a_2
S3=∑n=13an=a1+a2+a3 S_3 = \sum\limits_{n=1}^3 a_n = a_1 + a_2 + a_3

You now need to find, a1,a2 a_1, a_2 and a3 a_3 . Good luck, I hope what I've said helps.
(edited 7 years ago)

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