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The pattern goes like this: 0, 5, 12, 21, 32, 45

and I need to find out the formula for this pattern but I can't seem to figure it out.

Pls help me!

 Sep 21, 2017

Best Answer 

 #3
avatar+21850 
+1

The pattern goes like this: 0, 5, 12, 21, 32, 45

and I need to find out the formula for this pattern

 

\( \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 0} && 5 && 12 && 21 && 32 && 45 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 7 && 9 && 11 && 13 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 2} && 2 && 2 && 2 && \cdots \\ \end{array}\)

 

\(\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ \end{array}\)

 

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red} 0 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 2 } \\ &=& (n-1)\cdot 5 + \frac{n-1}{2}\cdot \frac{n-2}{1} \cdot 2 \\ &=& 5n - 5 + (n-1)(n-2) \\ &=& 5n - 5 + n^2-3n+2 \\ &=& 2n +n^2 -3 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{-3+2n+n^2} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_n} & \mathbf{=} & \mathbf{-3+2n+n^2} \\ a_1 &=& -3+2*1+1^2 \\ &=& -3+2+1 \\ &=& 0\ \checkmark \\\\ a_2 &=& -3 +2*2 + 2^2 \\ &=& -3+4+4 \\ &=& 5\ \checkmark \\\\ a_3 &=& -3 + 2*3 + 3^2 \\ &=& -3 + 6 + 9 \\ &=& 12\ \checkmark \\ \ldots \\ \hline \end{array}\)

.
 Sep 21, 2017
 #1
avatar+7372 
+1

0 + 5 = 5

5 + 7 = 12

12 + 9 = 21

21+ 11 = 32

32 + 13 = 45.

The pattern hides in the bold numbers :)

 Sep 21, 2017
 #2
avatar+70 
+1

I was answering the problem also Max, but realized he asked for an equation and didnt know how to do that? like (x+(5+(nth+2)))but that isnt correct

NediaMaster  Sep 21, 2017
edited by NediaMaster  Sep 21, 2017
 #3
avatar+21850 
+1
Best Answer

The pattern goes like this: 0, 5, 12, 21, 32, 45

and I need to find out the formula for this pattern

 

\( \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 0} && 5 && 12 && 21 && 32 && 45 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 7 && 9 && 11 && 13 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 2} && 2 && 2 && 2 && \cdots \\ \end{array}\)

 

\(\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ \end{array}\)

 

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red} 0 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 2 } \\ &=& (n-1)\cdot 5 + \frac{n-1}{2}\cdot \frac{n-2}{1} \cdot 2 \\ &=& 5n - 5 + (n-1)(n-2) \\ &=& 5n - 5 + n^2-3n+2 \\ &=& 2n +n^2 -3 \\ \mathbf{a_n} & \mathbf{=} & \mathbf{-3+2n+n^2} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_n} & \mathbf{=} & \mathbf{-3+2n+n^2} \\ a_1 &=& -3+2*1+1^2 \\ &=& -3+2+1 \\ &=& 0\ \checkmark \\\\ a_2 &=& -3 +2*2 + 2^2 \\ &=& -3+4+4 \\ &=& 5\ \checkmark \\\\ a_3 &=& -3 + 2*3 + 3^2 \\ &=& -3 + 6 + 9 \\ &=& 12\ \checkmark \\ \ldots \\ \hline \end{array}\)

heureka Sep 21, 2017

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