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Reason:
This is a geometric series with
a = -4 = first termr = -1/3 = common ratioThe template is [tex]a(r)^{n-1}[/tex]
If -1 < r < 1, then the infinite geometric series converges to a finite number. This is because we add on smaller and smaller pieces, which prevents the sum going off to infinity.
In the case of r = -1/3, it fits the interval -1 < r < 1. In other words -1 < -1/3 < 1 is true.
We'll plug those values into the formula below to wrap things up.
[tex]S = \frac{a}{1-r}\\\\S = \frac{-4}{1-(-1/3)}\\\\S = \frac{-4}{1+1/3}\\\\S = \frac{-4}{3/3+1/3}\\\\[/tex]
[tex]S = \frac{-4}{4/3}\\\\S = -4 \div \frac{4}{3}\\\\S = \frac{-4}{1} \times \frac{3}{4}\\\\S = \frac{-4*3}{1*4}\\\\S = -3\\\\[/tex]
Therefore,
[tex]\displaystyle \sum_{n=1}^{\infty} -4\left(-\frac{1}{3}\right)^{n-1} = -3[/tex]
The final answer is -3.
You can verify the answer by generating partial sums with a spreadsheet. The partial sums should steadily get closer to -3.
Here's a few partial sums.
[tex]\begin{array}{|c|c|c|} \cline{1-3}\text{n} & \text{a}_{\text{n}} & \text{S}_{\text{n}}\\\cline{1-3}1 & -4 & -4\\\cline{1-3}2 & 1.333333 & -2.666667\\\cline{1-3}3 & -0.444444 & -3.111111\\\cline{1-3}4 & 0.148148 & -2.962963\\\cline{1-3}5 & -0.049383 & -3.012346\\\cline{1-3}6 & 0.016461 & -2.995885\\\cline{1-3}7 & -0.005487 & -3.001372\\\cline{1-3}8 & 0.001829 & -2.999543\\\cline{1-3}9 & -0.00061 & -3.000153\\\cline{1-3}10 & 0.000203 & -2.99995\\\cline{1-3}\end{array}[/tex]
The interesting thing is that the partial sums [tex]S_n[/tex] bounce around -3 while also getting closer to it.
Please help! -screenshot-
A chemist wants to dilute a 30% phosphoric acid solution to a 15% solution. He needs 10 liters of the 15% solution. How many liters of the 30% solution and water must the chemist use?
The chemist needs to use 5 liters of the 30% solution and 5 liters of water.
Answer:
30% solution x liters
30%(water) 10-x liters
15% 10 liters
Step-by-step explanation: