11. Derive the formula for the sum of an arithmetic series
Let's call the number of terms [tex]n[/tex], the first term [tex]a_1[/tex] and constant difference [tex]d[/tex]
Each term in the series is [tex] a_k = a_1 + (k-1) d[/tex]
[tex]\displaystyle S_n = a_1 + a_2 + ... + a_n=\sum_{k=1}^n a_k =\sum_{k=1}^n(a_1 + (k-1) d)[/tex]
[tex]\displaystyle S_n = \sum_{k=1}^n a_1 + d\sum_{k=1}^n k - d \sum_{k=1}^n 1[/tex]
When we sum a constant n times, we're just going to get n times the constant. The sum of the first n natural numbers is n(n+1)/2 as Gauss knew at age eight.
[tex]S_n = n a_1 + dn(n+1)/2- dn[/tex]
[tex]S_n = n a_1 + \frac d 2 ( n^2 + n - 2n) = n (a_1 + \frac d 2 (n-1))[/tex]
That's a perfectly good formula, but we usually go further by noting
[tex]a_n = a_1 + (n-1)d[/tex]
[tex]S_n = \frac n 2 (a_1 + a_1 + d(n-1)) = n \cdot \dfrac{a_1 + a_n}{2}[/tex]
That's n times the average of the first and last element, which makes sense.
12.
[tex] \displaystyle \sum_{n=1}^5 (2+3n) = (2+3(1))+(2+3(2)) + (2+3(3)) + (2 + 3(4)) + (2 + 3(5)) [/tex]
[tex] \displaystyle \sum_{n=1}^5 (2+3n)=5 + 8 + 11 + 14 + 17[/tex]
13.
That's a common difference of 3, a first term of 20, and 12 terms
[tex]\displaystyle \sum_{k=1}^{12} ( 20 + 3(k-1)) [/tex]
14.
Common difference 5, first term 15, 14 terms
[tex]\displaystyle \sum_{k=1}^{14} ( 15 + 5(k-1)) [/tex]
