# Writing a recursive definition for Sn

• Feb 14th 2013, 12:57 PM
njuice8
Writing a recursive definition for Sn
Find a formula for tn, the nth term of the series. Then give a recursive definition for Sn, the sum of n terms of the series.

1) 8+12+18+27+....

2) 50+47+44+41+....

I know how to find the tn
tn = tn-1 * (3/2)
tn = 53-3n

My teacher gave us the answers
1. Sn = Sn-1 + 8(3/2)^(n-1)
2. Sn = Sn-1 +53-3n

Can somebody explain to me why these two are the answers for the recursive definition of Sn and how to write one? Thanks! (Happy)
• Feb 14th 2013, 01:39 PM
ILikeSerena
Re: Writing a recursive definition for Sn
Quote:

Originally Posted by njuice8
Find a formula for tn, the nth term of the series. Then give a recursive definition for Sn, the sum of n terms of the series.

1) 8+12+18+27+....

2) 50+47+44+41+....

I know how to find the tn
tn = tn-1 * (3/2)
tn = 53-3n

My teacher gave us the answers
1. Sn = Sn-1 + 8(3/2)^(n-1)
2. Sn = Sn-1 +53-3n

Can somebody explain to me why these two are the answers for the recursive definition of Sn and how to write one? Thanks! (Happy)

Hi njuice8! :)

A series is the sum of the terms.
So Sn is the sum of the first n terms.
Or put differently, the sum of the first (n-1) terms plus the nth term.
In formula form:

$S_n = t_1 + t_2 + ... + t_{n-1} + t_n$

$S_{n-1} = t_1 + t_2 + ... + t_{n-1}$

$S_n = S_{n-1} + t_n$

If you fill in the expression for $t_n$ you'll get both of the recursive formulas.
• Feb 16th 2013, 03:47 AM
ibdutt
Re: Writing a recursive definition for Sn
The first one is a GP with the first term a = 8 and common ration r = 3/2 { you are right}
n^th term will be given by ar^n-1 and sum of n terms Sn = ( r^n-1)/(r-1)
The second one is an AP with the first term a = 50 and common difference d = -3
The n^th term is given by Tn = a + ( n-1) d
and sum of the n terms Sn = n/2 [ 2a + ( n-1 ) d]
Plug in the values and get the answers.