What is the sum of #2/3+8/9+26/27+...# to #n# terms ?
1 Answer
Explanation:
The general term of a geometric series with initial term
#a_k = a*r^(k-1)#
The sum to
#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#
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
#sum_(k=1)^n a_k = (a(1-r^n))/(1-r)#