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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

Guest May 8, 2018
 #1
avatar+88898 
+2

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

 

 

cool cool cool

CPhill  May 8, 2018
 #2
avatar+20011 
0

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:

\(\begin{array}{|rcll|} \hline s_n &=& \left(\dfrac{a_1+a_n}{2}\right) \cdot n \quad & | \quad a_1 = -6 \qquad a_n = 10n-16 \\ &=& \left(\dfrac{-6+10n-16 }{2}\right) \cdot n \\ &=& \left(\dfrac{10n-22 }{2}\right) \cdot n \\ &=& (5n-11)\cdot n \\ \mathbf{s_n} & \mathbf{=} & \mathbf{5n^2-11n} \\ \hline \end{array} \)

 

laugh

heureka  May 9, 2018

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