An infinite geometric series has sum 2000. A new series, obtained by squaring each term of the original series, has sum 16 times the sum of the original series. The common ratio of the original series is m/n, where m and n are relatively prime positive integers. Find m+n.

Please help me.

DanielCai
Jul 12, 2018

#3**+3 **

Curious, different mistakes in #1 and #2 above, both leading to the same incorrect answer.

Suppose that the first series is

\(\displaystyle a, \ ar,\ ar^{2}, \ ar^{3}, \dots\)

Its sum to infinity is 2000, so

\(\displaystyle \frac{a}{1-r}=2000\dots(1)\)

For the second series, each term of the first series is squared, so the second series will be

\(\displaystyle a^{2}, \ a^{2}r^{2}, \ a^{2}r^{4}, \ a^{2}r^{6}, \dots\)

Its sum to infinity is 16 times the sum of the first series, so

\(\displaystyle \frac{a^{2}}{1-r^{2}}=32000\dots(2)\)

Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic

\(\displaystyle 63r^{2} - 125r +62 = 0,\)

from which

\(\displaystyle (63r-62)(r-1)=0,\)

so r = 62/63, (r can't equal 1).

Substitution in (1) shows that a = 2000/63.

Guest Jul 13, 2018

#1**+1 **

1-

Sum = F / [1 - R], where F= First term, R= Common ratio.

2000 = 16 / [1 - R], solve for R

R = 124 / 125

2-

32,000 = 256 / [1 - R], solve for R

R = 124 / 125

Therefore: m/n =124/125 and m + n = 124 + 125 =249

Guest Jul 12, 2018

#2**+1 **

2000 = F / [1 - R], ....................................(1)

32000 =F^2 / [1 - R]...................................(2)

From (1) above:

F =2000 - 2000R - Sub this for F in (2) above:

32000 =[2000 - 2000R]^2 / [1 -R]

Solve for R:

32000 = (2000 - 2000 R)^2/(1 - R)

32000 = (2000 - 2000 R)^2/(1 - R) is equivalent to (2000 - 2000 R)^2/(1 - R) = 32000:

(2000 - 2000 R)^2/(1 - R) = 32000

Multiply both sides by 1 - R:

(2000 - 2000 R)^2 = 32000 (1 - R)

Expand out terms of the right hand side:

(2000 - 2000 R)^2 = 32000 - 32000 R

Subtract 32000 - 32000 R from both sides:

-32000 + (2000 - 2000 R)^2 + 32000 R = 0

Expand out terms of the left hand side:

4000000 R^2 - 7968000 R + 3968000 = 0

The left hand side factors into a product with three terms:

32000 (R - 1) (125 R - 124) = 0

Divide both sides by 32000:

(R - 1) (125 R - 124) = 0

Split into two equations:

R - 1 = 0 or 125 R - 124 = 0

Add 1 to both sides:

R = 1 or 125 R - 124 = 0

Add 124 to both sides:

R = 1 or 125 R = 124

Divide both sides by 125:

R = 1 or R = 124/125

(2000 - 2000 R)^2/(1 - R) ⇒ (2000 - (2000 124)/125)^2/(1 - 124/125) = 32000:

So this solution is correct

(2000 - 2000 R)^2/(1 - R) ⇒ (2000 - 2000 1)^2/(1 - 1) = (undefined):

So this solution is incorrect

The solution is:

**R = 124/125**

Guest Jul 12, 2018

#3**+3 **

Best Answer

Curious, different mistakes in #1 and #2 above, both leading to the same incorrect answer.

Suppose that the first series is

\(\displaystyle a, \ ar,\ ar^{2}, \ ar^{3}, \dots\)

Its sum to infinity is 2000, so

\(\displaystyle \frac{a}{1-r}=2000\dots(1)\)

For the second series, each term of the first series is squared, so the second series will be

\(\displaystyle a^{2}, \ a^{2}r^{2}, \ a^{2}r^{4}, \ a^{2}r^{6}, \dots\)

Its sum to infinity is 16 times the sum of the first series, so

\(\displaystyle \frac{a^{2}}{1-r^{2}}=32000\dots(2)\)

Squaring equation (1) and using that to eliminate the a^2 term leads, after some simplification, to the quadratic

\(\displaystyle 63r^{2} - 125r +62 = 0,\)

from which

\(\displaystyle (63r-62)(r-1)=0,\)

so r = 62/63, (r can't equal 1).

Substitution in (1) shows that a = 2000/63.

Guest Jul 13, 2018