Answer:
0.1875
Step-by-step explanation:
The well formatted expression for the series has been attached to this response.
Each term in the series is govern by the rule, Tₙ
Where
n = term position
Tₙ = [tex]\frac{(-1)^{n+1}n}{2^n}[/tex]
To get the first six terms, we substitute n = 1 through 6 into Tₙ as follows:
When n = 1, we have;
T₁ = [tex]\frac{(-1)^{1+1}(1)}{2^1} = \frac{1}{2}[/tex] = 0.50000
When n = 2, we have;
T₂ = [tex]\frac{(-1)^{2+1}(2)}{2^2} = \frac{-2}{4} = \frac{-1}{2}[/tex] = -0.50000
When n = 3, we have;
T₃ = [tex]\frac{(-1)^{3+1}(3)}{2^3} = \frac{3}{8}[/tex] = 0.37500
When n = 4, we have;
T₄ = [tex]\frac{(-1)^{4+1}(4)}{2^4} = \frac{-1}{4}[/tex] = -0.25000
When n = 5, we have;
T₅ = [tex]\frac{(-1)^{5+1}(5)}{2^5} = \frac{5}{32}[/tex] = 0.15625
When n = 6, we have;
T₆ = [tex]\frac{(-1)^{6+1}(6)}{2^6} = \frac{-6}{64} = \frac{-3}{32}[/tex] = -0.09375
Therefore, the approximate sum of the series using the sum of the first six terms is
=> T₁ + T₂ + T₃ + T₄ + T₅ + T₆
=> 0.50000 + -0.50000 + 0.37500 + -0.25000 + 0.15625 + -0.09375
=> 0.1875
