Write both series as a summation in sigma notation.

Question

a). 3+2+5/3+3/2

b). 5+8+11+14+17+20

PLEASE HELP ME AND THANK YOU IF YOU DO!!!!

Answer 1

To write each series in sigma notation, we need to find the pattern in the terms.

Part a:

For part a, the series is:

3 + 2 + 5/3 + 3/2

Firstly, we can rewrite the whole numbers as fractions so the pattern is easier to recognize.

3 can be written as 3/1.

2 can be written as 4/2.

So, the series becomes:

3/1 + 4/2 + 5/3 + 6/4

Now, we can see the pattern. The denominators are 1, 2, 3, and 4. That means the denominator can be represented by n. The numerators are 3, 4, 5, 6. Each numerator is 2 more than the denominator. So, the numerator can be represented by n + 2. Therefore, the rule for each term is:

(n + 2) / n

Since there are 4 terms, n starts at 1 and ends at 4. So, for part a, it would be:

∑[(n + 2) / n] for n = 1 to n = 4

Part b:

For part b, the series is:

5 + 8 + 11 + 14 + 17 + 20

Firstly, look at how the terms change. From 5 to 8, 3 is being added. From 8 to 11, another 3 is being added and so on the pattern repeats. Since the same number is being added each time, this is an arithmetic sequence.

For an arithmetic sequence, the rule is:

a_n = a₁ + (n - 1)d

where,

a_n = term you are finding

a₁ = first term

n = term's position

d = common difference

So, we can substitute the first term and common difference:

5 + 3(n - 1)

Since there are 6 terms, n starts at 1 and ends at 6. So, the summation for part b would be:

∑[5 + 3(n - 1)] for n = 1 to n = 6

Answer 2

Sorry, I do not see the pattern in example a).

b however is the sum of an arithmetic series

∑[5+3(n-1)] for n=1 to 6