How do you find the sum of the infinite geometric series 12 + 6 + 3 + . . . ?

1 Answer
Nov 9, 2015

24

Explanation:

Let the series be

#S_1= sum_(i=1 to oo) a_i =12+6+3+ .... + a_n + ... +a_(oo)#

Consider the generic case: of #S_2#

Let k be any constant
Let r be the geometric ratio

Then #a_i = kr^i#

#color(blue)("Then "S_2 = kr^1 + kr^2 + kr^3 + ... + kr^n#......... ( 1 )

Note that you could have #" "kr^0 + kr^1 + kr^2 + ... + kr^n#

Multiply (1) by r giving:

#color(blue)( rS_2= kr^2 + kr^3 + kr^4 + ... +kr^n + kr^(n+1)#......... (2)

(2) - (1) ginves:

#color(brown)(rS_2 - S_2 =-kr^1+kr^(n+1) )#

#color(brown)(S_2 =( kr(r^n -1))/(r-1)#........( 3 )

Notice that the number sequence is reducing implying the #r<1#

.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find r consider #a_1/a_2 = (kr)/(kr^2) =1/r =12/6#

Inverting #1/r # gives #color(red)(r = 6/12 =1/2)#
.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find k
# kr - kr^2 = k(r-r^2) =k(1/2-1/4) =1/4k=12-6#

#1/4k = 6#

so #color(red)(k = 24)#
.~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Check")#
#S_2 = kr^1 + kr^2 + kr^3 -> (24 times 1/2) + (24 times 1/4) +( 24 times 1/8) .....#

#S_2 = 12+6+3+..#
this matches so it is safe to assume we have found the system.
.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Plug values for k and r in equation (3) and determine limits

#lim_(n -> oo) (24((1/2)^(n+1) -r))/(1/2-1)#

As #n -> oo , (1/2)^(n+1) -> 0 # so we end up with #-24r/(-(1/2) )= -24(1/2)/(-1/2)#

#color(red)(S_1 = +24)#