How do you find the sum of the finite geometric sequence of #Sigma (5/2)^(n-1)# from n=1 to 10?

1 Answer
May 24, 2018

Answer:

#sum_(n=1)^10 (5/2)^(n-1)=3254867/512=6357.162109375#

Explanation:

Solution

#sum_(n=1)^10 (5/2)^(n-1)#
#=sum_(m=0)^9 (5/2)^(m)# (you can still use #n# as the variable, but in effect you are letting #m=n-1#)
#=((5/2)^10-1)/((5/2)-1)# (geometric series formula)
#=3254867/512=6357.162109375#

Geometric Series Formula

From the series

#sum_(n=0)^x k^n = 1 + k + k^2 + ... + k^x#

We multiply throughout by k

#ksum_(n=0)^x k^n = k + k^2 + k^3 + ... + k^(x+1)#

Now subtract the original series from that sum

#ksum_(n=0)^x k^n - sum_(n=0)^x k^n = (k-1)sum_(n=0)^x k^n#

#= cancel(k) + cancel(k^2) + cancel(k^3) + ... + cancel(k^x) + k^(x+1) - (1 + cancel(k) + cancel(k^2) + ... + cancel(k^x)) #

#= k^(x+1) - 1#

So, dividing by (k-1), we have
#sum_(n=0)^x k^n = (k^(x+1) - 1)/(k-1)#