How do you find the explicit formula and calculate term 20 for 3, 9 , 27, 81, 243?

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How do you find the explicit formula and calculate term 20 for 3, 9 , 27, 81, 243?

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Answer 1 · 2015-07-17T18:33:00

The explicit formula for the progression is $t_{n} = 3^{n}$ $color(red)(t_n =3^n)$ and $t_{20} = 3 486 784 401$ $color(red)(t_20 = "3 486 784 401")$ .

Explanation:

This looks like a geometric sequence, so we first find the common ratio $r$ $r$ by dividing a term by its preceding term.

Your progression is $3, 9, 27, 81, 243$ $3, 9 , 27, 81, 243$ .

$\frac{t_{2}}{t_{1}} = \frac{9}{3} = 3$ $t_2/t_1 = 9/3= 3$

$\frac{t_{3}}{t_{2}} = \frac{27}{9} = 3$ $t_3/t_2 = 27/9= 3$

$\frac{t_{4}}{t_{3}} = \frac{81}{27} = 3$ $t_4/t_3 = 81/27= 3$

$\frac{t_{5}}{t_{4}} = \frac{243}{81} = 3$ $t_5/t_4 = 243/81 = 3$

So $r = 3$ $r = 3$ .

The $n^{th}$ $n^"th"$ term in a geometric progression is given by:

$t_{n} = a r^{n - 1}$ $t_n = ar^(n-1)$ where $a$ $a$ is the first term and $r$ $r$ is the common difference

So, for your progression.

$t_{n} = a r^{n - 1} = 3 {(3)}^{n - 1} = 3^{1} \times 3^{n - 1} = 3^{n - 1 + 1}$ $t_n = ar^(n-1) =3(3)^(n-1) = 3^1 × 3^(n-1) = 3^(n-1+1)$

$t_{n} = 3^{n}$ $t_n =3^n$

If $n = 20$ $n = 20$ , then

$t_{20} = 3^{20} = 3 486 784 401$ $t_20 = 3^20 = "3 486 784 401"$

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How do you find the explicit formula and calculate term 20 for 3, 9 , 27, 81, 243?

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