Question

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

Answer 1

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`

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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)`