Question

Can someone please help me with this...please

Answer 1

11. Derive the formula for the sum of an arithmetic series

Let's call the number of terms
`n`, the first term
`a_1` and constant difference
`d`

Each term in the series is
`a_k = a_1 + (k-1) d`

`\displaystyle S_n = a_1 + a_2 + ... + a_n=\sum_(k=1)^n a_k =\sum_(k=1)^n(a_1 + (k-1) d)`

`\displaystyle S_n = \sum_(k=1)^n a_1 + d\sum_(k=1)^n k - d \sum_(k=1)^n 1`

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.

`S_n = n a_1 + dn(n+1)/2- dn`

`S_n = n a_1 + \frac d 2 ( n^2 + n - 2n) = n (a_1 + \frac d 2 (n-1))`

That's a perfectly good formula, but we usually go further by noting

`a_n = a_1 + (n-1)d`

`S_n = \frac n 2 (a_1 + a_1 + d(n-1)) = n \cdot (a_1 + a_n)/(2)`

That's n times the average of the first and last element, which makes sense.

12.

`\displaystyle \sum_(n=1)^5 (2+3n) = (2+3(1))+(2+3(2)) + (2+3(3)) + (2 + 3(4)) + (2 + 3(5))`

`\displaystyle \sum_(n=1)^5 (2+3n)=5 + 8 + 11 + 14 + 17`

13.

That's a common difference of 3, a first term of 20, and 12 terms

`\displaystyle \sum_(k=1)^(12) ( 20 + 3(k-1))`

14.

Common difference 5, first term 15, 14 terms

`\displaystyle \sum_(k=1)^(14) ( 15 + 5(k-1))`