Respuesta :
Answer:
Please see the Step-by-step explanation for the answers
Step-by-step explanation:
1)
∑[tex]\left \ {{5} \atop {j=1}} \right.[/tex] 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:
∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] 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
∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] (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
∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] (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
∑[tex]\left \ {{4} \atop {k=1}} \right.[/tex] 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)
∑[tex]\left \ {{6} \atop {k=3}} \right.[/tex] (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:
[tex]ar^{n-1}[/tex] = [tex]1(\frac{1}{2})^{n-1}[/tex]
So the answer that represents the series in sigma notation is:
∑[tex]\left \ {{7} \atop {j=1}} \right.[/tex] [tex](\frac{1}{2})^{j-1}[/tex]
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:
∑[tex]\left \ {{5} \atop {j=1}} \right.[/tex] (2j−5)
For (1) the sum is 30, for (2) the sum is 90, for (3) the sum is -4, for(4) the sigma notation is [tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex] where j = 1 to j = 7, and for (5) the sigma notation is [tex]\rm\sum j = (2j-5)[/tex] where j = 1 to j = 5.
We have different series in the question.
It is required to find the sum of all series.
What is a series?
In mathematics, a series can be defined as a group of data that followed certain rules of arithmetic.
1) We have:
[tex]\rm \sum j=2j[/tex] where j = 1 to j = 5
After expanding the series, we get:
= 2(1)+2(2)+2(3)+2(4)+2(5)
=2(1+2+3+4+5)
= 2(15)
=30
2) We have:
[tex]\rm \sum k=(2k^2-4)[/tex] where k = 1 to k = 4
After expanding the series, we get:
[tex]\rm = (2(1)^2-4)+(2(2)^2-4)+(2(3)^2-4)+(2(4)^2-4)+(2(5)^2-4)\\[/tex]
[tex]\rm = 2[1^2+2^2+3^2+4^2+5^2]-4\times5\\\\\rm=2[55]-20\\\\\rm = 90[/tex]
3) We have:
[tex]\rm \sum k= (2k-10)[/tex] where k = 3 to k = 6
After expanding the series, we get:
[tex]= (2(3)-10)+(2(4)-10)+(2(5)-10)+(2(6)-10)\\\\=2[3+4+5+6] - 10\times4\\\\=2[18] - 40\\\\= -4[/tex]
4) The series given below:
[tex]1, \frac{1}{2} ,\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}[/tex]
It is a geometric progression:
[tex]\rm n^t^h[/tex] for the geometric progression is given by:
[tex]\rm a_n = ar^{n-1}[/tex]
[tex]\rm a_n = 1(\frac{1}{2})^{n-1}\\\\\rm a_n = (\frac{1}{2})^{n-1}\\[/tex]
In sigma notation we can write:
[tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex] where j = 1 to j = 7
5) The given series:
−3+(−1)+1+3+5, it is arithmetic series.
[tex]\rm n^t^h[/tex] for the arithmetic progression is given by:
[tex]\rm a_n = a+(n-1)d[/tex]
[tex]\rm a_n = -3+(n-1)(2)\\\\\rm a_n = 2n-5[/tex]
In sigma notation we can write:
[tex]\rm\sum j = (2j-5)[/tex] where j = 1 to j = 5
Thus, for (1) the sum is 30, for (2) the sum is 90, for (3) the sum is -4, for(4) the sigma notation is [tex]\rm \sum j = 1(\frac{1}{2})^{j-1}\\[/tex] where j = 1 to j = 7, and for (5) the sigma notation is [tex]\rm\sum j = (2j-5)[/tex] where j = 1 to j = 5.
Learn more about the series here:
https://brainly.com/question/10813422
