What is the sum of #2/3+8/9+26/27+...# to #n# terms ?

1 Answer
Apr 2, 2017

#sum_(k=1)^n a_k = n - 1/2 + 1/2(1/3)^n#

Explanation:

The general term of a geometric series with initial term #a# and common ratio #r# can be written:

#a_k = a*r^(k-1)#

The sum to #n# terms is given (see footnote) by the formula:

#sum_(k=1)^n = (a(1-r^n))/(1-r)#

Note that:

#2/3 = 1 - 1/3#

#8/9 = 1 - 1/9#

#26/27 = 1 - 1/27#

So it seems that the general term of the given series is:

#a_k = 1 - 1/3^k#

We have:

#sum_(k=1)^n 1/3^k = (1/3(1-(1/3)^n))/(1-1/3)#

#color(white)(sum_(k=1)^n 1/3^k) = (1-(1/3)^n)/2#

#color(white)(sum_(k=1)^n 1/3^k) = 1/2-1/2(1/3)^n#

So:

#sum_(k=1)^n a_k = sum_(k=1)^n (1-1/3^k)#

#color(white)(sum_(k=1)^n a_k) = sum_(k=1)^n 1 - sum_(k=1)^n 1/3^k#

#color(white)(sum_(k=1)^n a_k) = n - (1/2-1/2(1/3)^n)#

#color(white)(sum_(k=1)^n a_k) = n - 1/2 + 1/2(1/3)^n#

#color(white)()#
Footnote

Given:

#a_k = ar^(k-1)#

We find:

#(1-r)sum_(k=1)^n a_k = (1-r)sum_(k=1)^n ar^(k-1)#

#color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - rsum_(k=1)^n ar^(k-1)#

#color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - sum_(k=2)^(n+1) ar^(k-1)#

#color(white)((1-r)sum_(k=1)^n a_k) = a+color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - ar^n#

#color(white)((1-r)sum_(k=1)^n a_k) = a(1-r^n)#

So dividing both ends by #(1-r)# we get:

#sum_(k=1)^n a_k = (a(1-r^n))/(1-r)#