Find a general term an for the sequence whose first four terms are given. 2, 4, 8, 16, ...

Mathematics

1 answer:

lapo4ka [179]3 years ago

7 0

Answer:

an = 2·2^(n-1)

Step-by-step explanation:

There are simple tests to determine whether a sequence is arithmetic or geometric. The test for an arithmetic sequence is to check to see if the differences between terms are the same. Here the differences are 2, 4, 8, so are not the same.

The test for a geometric sequence is to check to see if the ratios of terms are the same. Here, the ratios are ...

4/2 = 2

8/4 = 2

16/8 = 2

These ratios are all the same (they are "common"), so the sequence is geometric.

The general term of a geometric sequence with first term a1 and common ratio r is ...

an = a1·r^(n-1)

Filling in the values for this sequence, we find the general term to be ...

an = 2·2^(n-1)

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Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.