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!
Hi njuice8!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!
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:
If you fill in the expression foryou'll get both of the recursive formulas.
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.
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