HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
TTTT
TTTH
TTHT
TTHH
THHH
THHT
THTH
THTT
Answer:
its A trust me
Step-by-step explanation:
Answer:
Given the series,
∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1
I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)
So,
∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞
There are different types of test to show if a series converges or diverges
So, using Ratio test
Lim n → ∞ (a_n+1 / a_n)
Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))
Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))
Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)
Lim n→ ∞ (-1/2)^(n-n+1)
Lim n→ ∞ (-1/2)^1 = -1/2
Since the limit is less than 0, then, the series converge...
Sum to infinity
Using geometric progression formula
S∞ = a / 1 - r
Where
a is first term
r is common ratio
So, first term is
a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1
a_1 = -4
Common ratio r = a_2 / a_1
a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2
a_2 = 2
Then,
r = a_2 / a_1 = 2 / -4 = -½
S∞ = -4 / 1--½
S∞ = -4 / 1 + ½
S∞ = -4 / 3/2 = -4 × 2 / 3
S∞ = -8 / 3 = -2⅔
The sum to infinity is -2.67 or -2⅔
<h2>
Step-by-step explanation: PHEW THAT TOOK A WHILE LOL IM A FAST TYPER</h2>
Answer:
Un alambre de acero de 1.5 m se estira 2.0 mm con fuerza F. El diámetro del alambre de acero es 4.0 mm. El módulo de Young del acero es 2.0 x 1011 Nm-2. Determine la fuerza F aplicada sobre el alambre.
Step-by-step explanation:
