Question

Urgent help! Need this to pass.

Answer 1

5)
The summation would be (A).
We need to compare the term to its value, the first term is 2, the second term is 4, the third term is 6.
We read this as:
2(1) + 2(2) + 2(3) + 2(4) + ... + 2(10)

The zero limit would mean it would start at 0 + 2 + 4, which is not what we wanted.

6)
Like above, we read the summation notation as:
5(3) + 5(4) + 5(5) + ... + 5(8) = 15 + 20 + 25 + 30 + 35 + 40 = 3(55) = 165

9)
Repeat as above, each term of n increases by 1 as we move from 1 to 10

12)
(a) Repeat as above.
(b) We can find whether it converges or diverges by finding the common ratio.
We do this by comparing the first two terms.

$T_1 = (-4)(\frac{1}{3})^{0} = -4$
$T_2 = (-4)(\frac{1}{3})^{1} = -4(\frac{1}{3})$

We can see that the ratio will be 1/3, which is less than 1.
This information tells us that the summation will converge, thus, we can find its sum.

(c) We find the sum by using the limiting sum formula.
$S_{\infty} = \frac{a}{1 - r}, \text{ }a = -4, \text{ }r = \frac{1}{3}$
$S_{\infty} = \frac{-4}{1 - \frac{1}{3}}$
$= \frac{-4}{\frac{2}{3}}$
$= -6$