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avatar+120 

Write recursive rule for sequence

1) 7, 20, 33, 46, 59.....

 

2) 27, 36, 48, 64, 256/2.....

 Nov 14, 2018
edited by BLANK  Nov 14, 2018
 #1
avatar+6099 
+1

\(1)~ a[n+1] = a[n]+13,~a[0]=7\)

.
 Nov 14, 2018
 #2
avatar+120 
+1

Is that all the steps of the process?

Wouldnt it need to be written in sigma notation for this kind of question?

or would u use the formula a[n+1] and that turns into a[n] +13 but i am confused.

where did that 13 come from?

and how dod you end up at a[0] = 7?

BLANK  Nov 14, 2018
 #3
avatar+6099 
+1

There's no magic here.  I noticed there is a difference of 13 between each number.

The first number given is 7.

Rom  Nov 14, 2018
edited by Rom  Nov 14, 2018
 #4
avatar+24093 
+12

Write recursive rule for sequence

 

1)

7, 20, 33, 46, 59.....

 

\(\begin{array}{|rcll|} \hline & \text{difference} \\ \hline 7 \\ & \color{red}13 \\ 20 \\ & \color{red}13 \\ 33 \\ & \color{red}13 \\ 46 \\ & \color{red}13 \\ 59 \\ \hline \end{array}\)

 

This is a Arithmetic Sequence

\(\text{Formula:}~ \boxed{a_n=a_1 + (n-1)d,\text{ with } a_1=7\text{ and }d = 13 }\)

\(\begin{array}{ll} \text{where:}\\ & a_1~ \text{ is the first term, and} \\ & d~ \text{ is the difference between the terms (called the "common difference")} \\ \end{array}\)

 

Write recursive rule

\(\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)d \\ a_{n+1} &=& a_1 + \Big((n+1)-1 \Big)d \\ \hline a_{n+1}-a_n &=& a_1 + \Big((n+1)-1 \Big)d - \Big( a_1 + (n-1)d \Big) \\ a_{n+1}-a_n &=& a_1 + nd - a_1 - (n-1)d \\ a_{n+1}-a_n &=& a_1 + nd - a_1 - nd +d \\ a_{n+1}-a_n &=& d \quad & | \quad +a_n \\ \mathbf{a_{n+1}} & \mathbf{=} & \mathbf{a_n+ d} \qquad d=\color{red}13 \\ \mathbf{a_{n+1}} & \mathbf{=} & \mathbf{a_n+\color{red}13} \\ \hline \end{array} \)

 

laugh

 Nov 14, 2018
 #5
avatar+6099 
+1

second one is simple when you see it.

 

\(a[n+1] = \dfrac 4 3 a[n],~a[0]=27\)

 

This does end up with  \(\dfrac{256}{3} \text{ instead of }\dfrac{256}{2}\)

 

I'm going to assume yours is a typo.

 Nov 14, 2018

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