given that the following two are geometric series are convergent: 1+x+x^2+x^3+...and 1-x+x^2-x^3+... determine the value(s) of x for which the sum of the two series is equal to 8​

Let S and T denote the two finite sums,

S = 1 + x + x ² + x ³ + … + x ᴺ

T = 1 - x + x ² - x ³ + … + (-x) ᴺ

• If both S = 8 and T = 8 as N goes to infinity:

Then

xS = x + x ² + x ³ + x ⁴ + … + x ᴺ⁺¹

-xT = -x + x ² - x ³ + x ⁴ + … + (-x) ᴺ⁺¹

so that

S - xS = 1 - x ᴺ⁺¹   ==>   S = (1 - x ᴺ⁺¹)/(1 - x)

and similarly,

T = (1 - (-x) ᴺ⁺¹)/(1 + x)

For both sums, so long as |x| < 1, we have

lim [N → ∞] S = 1/(1 - x)

lim [N → ∞] T = 1/(1 + x)

Then if both sums converge to 8, this happens for

S : 1/(1 - x) = 8   ==>   x = 7/8

T : 1/(1 + x) = 8   ==>   x = -7/8

• If the sum S + T = 8 as N goes to infinity:

From the previous results, we have

1/(1 - x) + 1/(1 + x) = 8   ==>   x = ±√3/2

answer: 457.14

step-by-step explanation:

205 - 45 = 160

160 / .35 = 457.142857