The Punic Wars began in 264 B.C. and ended in 146 B.C. How long did the Punic Wars last??
2 answers:
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Answer:
Please see the Step-by-step explanation for the answers
Step-by-step explanation:
1)
∑ 2j
The sum of series from j=1 to j=5 is:
∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)
= 2 + 4 + 6 + 8 + 10
∑ = 30
2)
This question is not given clearly so i assume the following series that will give you an idea how to solve this:
∑ 2k²
The sum of series from k=1 to j=4 is:
∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²
= 2(1) + 2(4) + 2(9) + 2(16)
= 2 + 8 + 18 + 32
∑ = 60
∑ (2k)²
∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²
= (2)² + (4)² + (6)² + (8)²
= 4 + 16 + 36 + 64
∑ = 120
∑ (2k)²- 4
∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4
= (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4
= (4-4) + (16-4) + (36-4) + (64-4)
= 0 + 12 + 32 + 60
∑ = 104
∑ 2k²- 4
∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4
= 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4
= (2-4) + (8-4) + (18-4) + (32-4)
= -2 + 4 + 14 + 28
∑ = 44
3)
∑ (2k-10)
∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)
= (6-10) + (8-10) + (10-10) + (12-10)
= -4 + -2 + 0 + 2
∑ = -4
4)
1+1/2+1/4+1/8+1/16+1/32+1/64
This is a geometric sequence where first term is 1 and the common ratio is 1/2 So
a = 1
This can be derived as
1/2/1 = 1/2 * 1 = 1/2
1/4/1/2 = 1/4 * 2/1 = 1/2
1/8/1/4 = 1/8 * 4/1 = 1/2
1/16/1/8 = 1/16 * 8/1 = 1/2
1/32/1/16 = 1/32 * 16/1 = 1/2
1/64/1/32 = 1/64 * 32/1 = 1/2
Hence the common ratio is r = 1/2
So n-th term is:
=
So the answer that represents the series in sigma notation is:
∑
5)
−3+(−1)+1+3+5
This is an arithmetic sequence where the first term is -3 and the common difference is 2. So
a = 1
This can be derived as
-1 - (-3) = -1 + 3 = 2
1 - (-1) = 1 + 1 = 2
3 - 1 = 2
5 - 3 = 2
Hence the common difference d = 2
The nth term is:
a + (n - 1) d
= -3 + (n−1)2
= -3 + 2(n−1)
= -3 + 2n - 2
= 2n - 5
So the answer that represents the series in sigma notation is:
∑ (2j−5)