evaluate each series described.
9) $\sum_{m = 1}^{6} (10m - 3)$
10) $\sum_{n = 1}^{7} (14 - 4n)$
11) $\sum_{k = 1}^{10} 4^{k - 1}$
12) $\sum_{i = 1}^{8} (-6)^{i - 1}$
given the recursive formula for the sequence, find the first five terms.
13) $a_n = a_{n - 1} + 9$, $a_1 = 16$
14) $a_n = a_{n - 1} + 8$, $a_1 = -28$
15) $a_n = a_{n - 1} - 2$, $a_1 = -30$
16) $a_n = a_{n - 1} - 6$, $a_1 = 39$
17) $a_n = a_{n - 1} \cdot 5$, $a_1 = 2$
18) $a_n = a_{n - 1} \cdot -2$, $a_1 = 3$
19) $a_n = a_{n - 1} \cdot 4$, $a_1 = -2$
20) $a_n = a_{n - 1} \cdot -5$, $a_1 = 2$

Question

Accepted Answer

Let's solve problem 9: $\sum_{m = 1}^{6}(10m - 3)$

# Explanation:
## Step 1: Expand the summation
We need to find the sum of the series by substituting $m = 1, 2, 3, 4, 5, 6$ into the expression $10m - 3$ and then adding them up.

For $m = 1$: $10(1) - 3 = 10 - 3 = 7$

For $m = 2$: $10(2) - 3 = 20 - 3 = 17$

For $m = 3$: $10(3) - 3 = 30 - 3 = 27$

For $m = 4$: $10(4) - 3 = 40 - 3 = 37$

For $m = 5$: $10(5) - 3 = 50 - 3 = 47$

For $m = 6$: $10(6) - 3 = 60 - 3 = 57$

## Step 2: Sum the terms
Now we add all these terms together: $7 + 17 + 27 + 37 + 47 + 57$

We can use the formula for the sum of an arithmetic series $S_n=\frac{n(a_1 + a_n)}{2}$, where $n = 6$ (number of terms), $a_1 = 7$ (first term), and $a_n = 57$ (last term).

$S_6=\frac{6(7 + 57)}{2}=\frac{6	imes64}{2}= 3	imes64 = 192$

# Answer:
192

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QUESTION IMAGE

Question

Explanation:

Step 1: Expand the summation

Step 2: Sum the terms

Answer: