What is the common ratio of the series #50+40+38...#?
I know how to find the sum of #S_(oo)# when #r# is obvious, like with #sum_(n=1)^(oo)(1/2^n)# , but I don't know how to find the sum when I don't know the common ratio.
I know how to find the sum of
2 Answers
The given series is not a Geom. Series, and, as such, we
can not have the Common Ratio.
Explanation:
The Common Ratio, associated with the following
Geometric Series, is
Its General
We note that,
As far as the given series is concerned, we find that,
So, the given series is not a Geom. Series, and, as such, we
can not have the Common Ratio.
The terms fit a quadratic sequence of the form:
Explanation:
We are only given 3 terms which limits the accuracy of any conclusion drawn. Let us assume that it is a quadratic sequence. Without additional terms we can be certain this is the correct assumption.
Take the difference between consecutive terms, and then take the difference between those terms, as follows:
# {: ("Seq: ",50,,40,,38), ("1st Difference:",,-10,,-2,),("2nd difference:",,,8,,) :} #
So, if this is a quadratic sequence, the terms form a quadratic sequence, of the form:
# {u_n} = an^2 + bx + c #
and the quadratic coefficient is twice the 2nd difference, ie:
# 2a=8 => a=4 #
We now form a table where we take the original sequence and subtract the quadratic term to give a residue:
# {: (n:, 1,2,3), (n^2:, 1,4,9), ("Seq ("n"): ", 50,40,38), (an^2: ,4,16,36), ("Residue:" ,46,24,2) :}#
So we can now also conclude that the linear part of the sequence (
# \ \ \ \ \ u_n = an^2+bn+c #
# :. u_n =4n^2-22n+68 #
Which we can verify as follows:
# {: (n:, 1,2,3), (n^2: ,1,4,9), (ul(" ") ,ul(" "),ul(" "),ul(" ")),(4n^2 ,4,16,36),(-22n ,-22,-44,-66),(+68 ,68,68,68),(ul(" ") ,ul(" "),ul(" "),ul(" ")),(u_n ,50,40,38) :}#