Respuesta :
Notice that
a + ar³ + ar⁶ + ar⁹ + ... = a + r³ (a + ar³ + ar⁶ + ...)
but the grouped sum on the right side is the same as the sum on the left side. This means that, if the sum we want to compute
a + ar³ + ar⁶ + ar⁹ + ...
converges to S, then
S = a + r³ S
Solve for S :
S - r³ S = a
(1 - r³) S = a
S = a/(1 - r³)
Similarly, knowing the value of the first sum, we have
a + ar + ar² + ar³ + ... = a + r (a + ar + ar² + ...)
but we know
a + ar + ar² + ... = 15
so
15 = a + 15r
and knowing the value of the second sum, we have
a + ar² + ar⁴ + ... = a + r² (a + ar² + ar⁴ + ...)
so
9 = a + 9r²
Solve these two equations for a and r :
15 = a + 15r ⇒ a = 15 - 15r
9 = a + 9r² ⇒ 9 = 15 - 15r + 9r² ⇒ 3r² - 5r + 2 = (3r - 2) (r - 1) = 0
⇒ r = 2/3 or r = 1
We cannot have r = 1, since that would mean
a + a + a + ... = 15
since the left side is adding infinitely many copies of some constant; such a sum can never converge to a finite number. So it must be that r = 2/3, and it follows that a = 15 - 15 (2/3) = 5.
Then the sum we want has a value of
S = 5/(1 - (2/3))³ = 135/19
Answer:
Step-by-step explanation:
a + ar¹ + ar² + ar³ + ... = 15 #1
r(a + ar¹ + ar² + ar³ + ...) = 15r
ar¹ + ar² + ar³ + ... = 15r #2
subtract #2 from #1
a = 15 - 15r
a + ar² + ar⁴ + ar⁶ + ... = 9 #3
r²(a + ar² + ar⁴ + ar⁶ + ... )= 9r²
ar² + ar⁴ + ar⁶ + ... = 9r² #4
subtract #4 from #3
a = 9 - 9r²
15 - 15r = 9 - 9r²
9r² - 15r + 6 = 0
3r² - 5r + 2 = 0
r = (5 ±√(5² - 4(3)(2))) / (2(3)
r = (5 ± 1) / 6
r = 1
or
r = 4/6 = 2/3
if r = 1 then a = 15 - 15(1) = 0 and a series of zeros added together can never equal 15 so r = 2/3 must be true.
a = 15 - 15(2/3) = 5
a + ar³ + ar⁶ + ar⁹ + ... = C #5
r³(a + ar³ + ar⁶ + ar⁹ + ...) = Cr³
ar³ + ar⁶ + ar⁹ + ... = Cr³ #6
subtract #6 from #5
a = C - Cr³
a = C(1 - r³)
C = a/(1 - r³)
C = 5/(1 - (2/3)³)
C = 5/(1 - 8/27)
C = 5/(19/27)
C = 5(27/19)
C = 135/19 = 7.1052631578947368....
