Expert: Paul Klarreich - 9/26/2009

Question
(e) the sum to infinity of a G.P series is R. The sum to infinity of the squares of the terms is 2R. The sum to infinity of the cubes of the terms is (64/13)R. Find (i) the value of R. (ii) the first term of the first original series.


Answer
Questioner: Edwin
Country: Norway
Category: Advanced Math
Private: No
Subject: Math
Question: (e) the sum to infinity of a G.P series is R. The sum to infinity of the squares of the terms is 2R. The sum to infinity of the cubes of the terms is (64/13)R. Find (i) the value of R. (ii) the first term of the first original series.
---------------------------------------
First sentence:

a + ar + ar^2 + .... =

 a
------ = R,
1 - r

a = R(1 - r)
.................................
Second sentence:

a^2 + (ar)^2 + (ar^2)^2 + ....

= a^2(1 + r^2 + r^4 + ... ) =
    a^2
= ---------- = 2R
  1 - r^2

a^2 = 2R(1 - r^2)
a = R(1 - r)
--------------
a = 2(1 + r)

R(1 - r) = 2(1 + r)
    2(1 + r)
R = ---------
     1 - r

......................
Third sentence:
a^3 + (ar)^3 + .. + (ar^k)^3 + ....

= a^3(1 + r^3 + r^6 + ... ) = 64R/13
  a^3
---------- = 64R/13
1 - r^3

...................................
Now we have three equations in three variables, a,r,R.
  a^2               2a
---------- = 2R =  ------
1 - r^2           1 - r

  a        
---------- = 2   << divide out a.
1 + r    

a = 2(1 + r)    << a in terms of r

    2(1 + r)
R = ---------  << R in terms of r
     1 - r
Use this, and substitute for  a and R:
  a^3
---------- = 64R/13
1 - r^3

8(1 + r)^3   64[2(1 + r)]
---------- = -------------
1 - r^3      13(1 - r)


(1 + r)^2        8 [2]
---------- = -------------   << a little canceling
1 - r^3      13(1 - r)


(1 + r)^2                16
------------------- = -------------   << a little factoring
(1-r)(1 + r + r^2)     13(1 - r)

1 + 2r + r^2         16
---------------- = ------   << a little canceling
1 + r + r^2          13


13 + 26r + 13r^2 = 16 +  16r + 16r^2  << cross-multiply

0 = 3 - 10r + 3r^2  << simplify

0 = (3 - r)(1 - 3r)  << factor

r = 3,   r = 1/3    << solve.

We cannot have  r = 3, because the series would diverge.

So  r = 1/3

a = 2(1 + r) = 2(4/3) = 8/3

    2(4/3)
R = ------- = 4
     2/3

...........................
There is your  a,r,R, now we check:

                             8/3      8/3   
8/3(1 + 1/3 + 1/9 + ...) =  ------- = ----- = 4  = R (good)
                           1 - 1/3    2/3


                             64/9        64/9
64/9(1 + 1/9 + 1/81 + ...) = -------- = ------- = 8 = 2R  (really good)
                            1 - 1/9      8/9


512/27(1 + 1/27 + 1/729 + ..) =

512/27     512/27    256    64(4)   64R
-------- = -------- = ---- = ----- = ----  (really, really good)
1 - 1/27     26/27     13     13      13