Question

Solve for 50 points

[ a ( 1/r + 1 + r ) = 21]^2

Answer 1

Answer:

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + ...

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + ...

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + ...

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + ...

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + ... + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + ... + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, ..., a r^n all cancel and you are left with S - r S = a - a r^(n+1), so n+1 a (1 - r ) S = ------------ . 1 - r

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

Answer 2

Step-by-step explanation:

(a(1/r+1+r)=21)^2

(a(1+r+2r÷r)=21)^2

(a(1+3r)=21×r)^2

(a+3ra=21r)^2

a^2+ 324r^2