Question

How do you find the explicit formula for the following sequence $3,6,12,24,48$?

Answer 1

Hint: From the given series of geometric sequences, we find the general term of the series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. From the given sequence we find the common ratio which is the ratio between two consecutive terms. We put the values to get the formula for the general term ${{t}_{n}}$. Then we put the value of consecutive natural numbers for $n$ to find the solution.

Complete answer:
We have been given a series of geometric sequence which is $3,6,12,24,48......$
We express the geometric sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common ratio be $r$ where $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common ratio.
The formula being ${{t}_{n}}={{t}_{1}}{{r}^{n-1}}$.
The first term is 3. So, ${{t}_{1}}=3$. The common difference is $r=\dfrac{6}{3}=\dfrac{12}{6}=\dfrac{24}{12}=2$.
We put the values of ${{t}_{1}}$ and $r$ to find the general form.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}{{r}^{n-1}}=3\times {{2}^{n-1}}$.
Now we place consecutive natural numbers for $n$ as $1,2,3,4,...$ to get the sequence as
\[3\times {{2}^{1-1}},3\times {{2}^{2-1}},3\times {{2}^{3-1}},3\times {{2}^{4-1}},3\times {{2}^{5-1}},3\times {{2}^{6-1}}......=3,6,12,24,48,96......\] .

Note: The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number. The ratio formula should always be according $r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}}$.