[tex]11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i[/tex]
Solution:
Given that,
11 + 17 + 23 + 29 + 35 + 41
We have to express the sum in sigma notation
Analyse the series
5 + 6(1) = 11
5 + 6(2) = 17
5 + 6(3) = 5 + 18 = 23
5 + 6(4) = 5 + 24 = 29
5 + 6(5) = 5 + 30 = 35
5 + 6(6) = 5 + 36 = 41
Thus the series goes on like this:
5 + 6(n) , where n = 1 to 6
This can be expressed in sigma notation as:
[tex]11+17+23+29+35+41=\sum\limits_{n=1}^{6}5+6n[/tex]
We can expand the sigma notation and verify the results
[tex]\sum\limits_{i=1}^{6}5+6i=(5+6(1))+(5+6(2))+(5+6(3))+(5+6(4))+(5+6(5))+(5+6(6))\\\\\\\sum\limits_{i=1}^{6}5+6i=(5+6)+(5+12)+(5+18)+(5+24)+(5+30)+(5+36)\\\\\\\sum\limits_{i=1}^{6}5+6i=11+17+23+29+35+41[/tex]
The given sum is expressed by the notation, ∑ 11 + 6n, where the lower limit is n = 0 and the upper limit is n =5.
Step-by-step explanation:
Step 1; First, we need to determine how the numbers in the series (11 + 17 + 23 + 29 + 35 + 41) are related to one another. The initial value is 11 and for every successive number, there is a difference of 6.
First number = 11,
Second number = 11 + 6 = 11 + (6) × 1 = 17,
Third number = 11 + 6 + 6 = 11 + (6) × 2 = 23,
Fourth number = 11 + 6 + 6 + 6 = 11 + (6) × 3 = 29,
Fifth number = 11 + 6 + 6 + 6 + 6 = 11 + (6) × 4 = 35,
Sixth number = 11 + 6 + 6 + 6 + 6 + 6 = 11 + (6) × 5 = 41.
Step 2; So for the sigma notation, we insert the initial term i.e. 11 and to represent the term that is added, we put 6n.
So by substituting the values in the notation, ∑ 11 + 6n, where the lower limit is n = 0 and upper limit is n =5, we get;
n = 0, 11 + 6n = 11 + (6) × 0 = 11,
n = 1, 11 + 6n = 11 + (6) × 1 = 17,
n = 2, 11 + 6n = 11 + (6) × 2 = 23,
n = 3, 11 + 6n = 11 + (6) × 3 = 29,
n = 4, 11 + 6n = 11 + (6) × 4 = 35,
n = 5, 11 + 6n = 11 + (6) × 5 = 41.
The value of ∑ for all these values is given by 11 + 17 + 23 + 29 + 35 + 41 = 156.
