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is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

 Jul 8, 2015

Best Answer 

 #2
avatar+22892 
+10

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

 

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{$
\begin{array}{lcccccccccc}
$Number $a &a_1=\textcolor[rgb]{1,0,0}{ 14}& & 21& & 42& &77 & &126 & \cdots \\
$First difference $D^1 & & D_0^1=\textcolor[rgb]{1,0,0}{7}& & 21 & & 35 & & 49 & \cdots \\
$Second difference $D^2 & & & D_0^2=\textcolor[rgb]{1,0,0}{14}& & 14& &14 & \cdots \\
\end{array}
$}}$$

 

If we have a arithmetic sequence of order k, we can find $$a_n$$ by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k
~~}$$

So the nth term is given by:

$$\small{\text{$
\begin{array}{rcl}
a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + \binom{n-1}{1}\cdot \textcolor[rgb]{1,0,0}{7} + \binom{n-1}{2}\cdot \textcolor[rgb]{1,0,0}{14} \qquad
| \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + \dfrac{(n-2)(n-1)}{2}\cdot \textcolor[rgb]{1,0,0}{14} \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + (n-2)(n-1)\cdot 7 \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)[1+(n-2)]\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)(n-1)\\\\
\mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\
\end{array}
$}}$$

 

 Jul 9, 2015
 #1
avatar+102386 
+5

This series is not arithmetic, because there is no common difference ......the nth term - for n ≥ 2 - is given by:

 

14 + 7(n-1)^2  

 

 

      

 Jul 8, 2015
 #2
avatar+22892 
+10
Best Answer

is the sequence arithmetic?if so, identify the common difference. 14,21,42,77...

 

This ist a ARITHMETIC SEQUENCE OF HIGHER ORDER.

The sequence  is arithmetic of order k if the differences of order k are equal.

We have the order k = 2. The second differences are equal = 14.

Let us see:

$$\small{\text{$
\begin{array}{lcccccccccc}
$Number $a &a_1=\textcolor[rgb]{1,0,0}{ 14}& & 21& & 42& &77 & &126 & \cdots \\
$First difference $D^1 & & D_0^1=\textcolor[rgb]{1,0,0}{7}& & 21 & & 35 & & 49 & \cdots \\
$Second difference $D^2 & & & D_0^2=\textcolor[rgb]{1,0,0}{14}& & 14& &14 & \cdots \\
\end{array}
$}}$$

 

If we have a arithmetic sequence of order k, we can find $$a_n$$ by :

$$\boxed{~~a_n = a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2+\cdots+\binom{n-1}{k}\cdot D_0^k
~~}$$

So the nth term is given by:

$$\small{\text{$
\begin{array}{rcl}
a_n &=& a_1 + \binom{n-1}{1}\cdot D_0^1 + \binom{n-1}{2}\cdot D_0^2\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + \binom{n-1}{1}\cdot \textcolor[rgb]{1,0,0}{7} + \binom{n-1}{2}\cdot \textcolor[rgb]{1,0,0}{14} \qquad
| \qquad \binom{n-1}{1} = n-1 \qquad \binom{n-1}{2}=\dfrac{(n-2)(n-1)}{2}\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + \dfrac{(n-2)(n-1)}{2}\cdot \textcolor[rgb]{1,0,0}{14} \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + (n-1)\cdot \textcolor[rgb]{1,0,0}{7} + (n-2)(n-1)\cdot 7 \\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)[1+(n-2)]\\\\
a_n &=& \textcolor[rgb]{1,0,0}{14} + 7(n-1)(n-1)\\\\
\mathbf{a_n} & \mathbf{=} & \mathbf{14 + 7(n-1)^2} \qquad | \qquad n \ge 1 \\
\end{array}
$}}$$

 

heureka Jul 9, 2015

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