Question

How do you find a formula of the nth term of the sequence $\text{1, 3, 9, 27, 81, }...$ ?

Answer 1

Hint: In this question, we have to find the formula for the nth term of a sequence. A sequence is a collection of objects or numbers, which is represented in order and repetitions are also allowed. So, we see that the given sequence is a type of geometric sequence because we see that the first term obtained is the multiplication of the previous term by a fixed number. So, we will solve this problem by finding a common pattern between the ratio and then put the first term and common ratio in the nth form of the geometric sequence, to get the required solution to the problem.

Complete step-by-step answer:
According to the question, we have to find the nth term formula of the sequence.
So, we will find the pattern in the given sequence.
The sequence given to us is $\text{1, 3, 9, 27, 81, }...$ ------------- (1)
So, the first term of the sequence is represented as ${{a}_{1}}$ , therefore from equation (), we get${{a}_{1}}=1$ , and the remaining terms of the sequence with their values are represented as ${{a}_{1}}=1,{{a}_{2}}=3,{{a}_{3}}=9$ .
Now, we will see that any two consecutive terms of the given sequence have a common difference or a common ratio. So, let us first see the difference:
$d={{a}_{2}}-{{a}_{1}}=3-1=2$ and
$d={{a}_{3}}-{{a}_{2}}=9-3=6$
Thus, we see that in the above two equations, the difference between two consecutive numbers is not the same. Therefore, we now find the common ratio of the sequence, that is
$r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{3}{1}=3$ and
$r=\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{9}{3}=3$
Thus, we see that the common ratio of the two consecutive numbers is the same, therefore the type of sequence is a geometric sequence.
Therefore, we see a pattern in the sequence, that is
$\Rightarrow r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=....=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
$\Rightarrow r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
So, now we will multiply ${{a}_{n-1}}$ on both sides of the equation, we get
$\Rightarrow r.{{a}_{n-1}}=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}.({{a}_{n-1}})$
On further solving, we get
$\Rightarrow r.{{a}_{n-1}}={{a}_{n}}$
Now, we know ${{a}_{n-1}}=r.{{a}_{n-2}}$ , so put this value in the above equation, we get
 $\begin{align}
  & \Rightarrow r.r.{{a}_{n-2}}={{a}_{n}} \\
 & \Rightarrow {{r}^{2}}{{a}_{n-2}}={{a}_{n}} \\
\end{align}$
Again if we substitute the value ${{a}_{n-2}}=r.{{a}_{n-3}}$ in the above equation, we get
$\begin{align}
  & \Rightarrow r.r.r.{{a}_{n-3}}={{a}_{n}} \\
 & \Rightarrow {{r}^{3}}{{a}_{n-3}}={{a}_{n}} \\
\end{align}$
Thus, in the end, we get
$\Rightarrow {{r}^{n-1}}a={{a}_{n}}$
Therefore, the general nth form of the geometric sequence is ${{r}^{n-1}}a={{a}_{n}}$ .
So, now we will put the value of the first term and the common ratio in the above formula, we get
$\Rightarrow {{3}^{n-1}}.(1)={{a}_{n}}$
Thus, on further solving we get
$\Rightarrow {{a}_{n}}={{3}^{(n-1)}}$
Therefore, for the sequence $\text{1, 3, 9, 27, 81, }...$ , the general formula for nth terms is equal to ${{r}^{n-1}}a={{a}_{n}}$ and the nth term of the sequence is ${{3}^{\left( n-1 \right)}}$ .

Note: While solving this problem, do mention the idea you are using to avoid errors and calculation mistakes. Do not confuse the geometric sequence and the arithmetic sequence. First, do find whether the given sequence has a ratio or the difference between any two consecutive numbers. In addition, find the formula to get the required solution to the problem.