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)

Re: Writing a recursive definition for Sn

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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:

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

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

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

If you fill in the expression for $\displaystyle t_n$ you'll get both of the recursive formulas.

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.