Consider the arithmetic series:
-6 + 4 + 14 + 24 + ...
Write a formula for the sum of the first n terms in this series.
10n + 10
10n + 4
5n2−11n
5n2-11
The first few terms are
-6 -2 12 36 70
And using the sum of differences, we have :
-6 -2 12 36 70
4 14 24 34
10 10 10
This will be a 2nd power polynomial in the form an^2 + bn + c
And we have this system
a + b + c = -6
4a + 2b + c = -2
9a + 3b + c = 12
Subtract the frist equation from the second and the second equation from the third and we have
3a + b = 4
5a + b = 14 subtract the first equation from the second
2a = 10
a = 5
And using 3a + b = 4 ⇒ 3(5) + b = 4 ⇒ b = -11
And using a + b + c = -6 ⇒ 5 - 11 + c = -6 ⇒ -6 + c = -6 ⇒ c = 0
So....the polynomial is
5n2 - 11n
Consider the arithmetic series:
-6 + 4 + 14 + 24 + ...
Write a formula for the sum of the first n terms in this series.
10n + 10
10n + 4
5n2−11n
5n2-11
\(\begin{array}{lrrrrrrrrrr} & -6 && 4 && 14 && 24 && \cdots \\ \text{1. Difference } && 10 && 10 && 10 && \cdots \\ \end{array} \)
Formula arithmetic series:
\(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)d \quad & | \quad a_1 = -6 \qquad d = 10 \\ a_n &=& -6 + (n-1)10 \\ a_n &=& -6 + 10n-10 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{10n-16} \\ \hline \end{array} \)
Formula the sum of a arithmetic series:
\(\begin{array}{|rcll|} \hline s_n &=& \left(\dfrac{a_1+a_n}{2}\right) \cdot n \quad & | \quad a_1 = -6 \qquad a_n = 10n-16 \\ &=& \left(\dfrac{-6+10n-16 }{2}\right) \cdot n \\ &=& \left(\dfrac{10n-22 }{2}\right) \cdot n \\ &=& (5n-11)\cdot n \\ \mathbf{s_n} & \mathbf{=} & \mathbf{5n^2-11n} \\ \hline \end{array} \)