Find nth term of series 3, 11, 31, 69, . . . . .
Given an integer N, the task is to find the Nth term of the series
3, 11, 31, 69, . . . . . till Nth term.
Examples:
Input: N = 3
Output: 31
Input: N = 6
Output: 223
Approach:
From the given series, find the formula for Nth term–
1st term = 1 ^ 3 + (1 + 1) = 3
2nd term = 2 ^ 3 + (2 + 1) = 11
3rd term = 3 ^ 3 + (3 + 1) = 31
4th term = 4 ^ 3 + (4 + 1) = 69
.
.
Nth term = n ^ 3 + (n + 1)
The Nth term of the given series can be generalized as-
TN = n ^ 3 + (n + 1)
Illustration:
Input: N = 5
Output: 131
Explanation:
TN = n ^ 3 + (n + 1)
= 5 ^ 3 + (5 + 1)
= 131
Below is the implementation of the above approach-
C++
#include <iostream>
using namespace std;
int find_nth_Term( int n)
{
return n * n * n + (n + 1);
}
int main()
{
int n = 5;
cout << find_nth_Term(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int find_nth_Term( int n)
{
return n * n * n + (n + 1 );
}
public static void main(String[] args)
{
int n = 5 ;
System.out.println(find_nth_Term(n));
}
}
|
Python
def find_nth_Term(n):
return n * n * n + (n + 1 )
n = 5
print (find_nth_Term(n))
|
C#
using System;
class GFG
{
static int find_nth_Term( int n)
{
return n * n * n + (n + 1);
}
public static int Main()
{
int n = 5;
Console.WriteLine(find_nth_Term(n));
return 0;
}
}
|
Javascript
<script>
function find_nth_Term(n)
{
return n * n * n + (n + 1);
}
let n = 5;
document.write(find_nth_Term(n))
</script>
|
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
Last Updated :
20 Aug, 2022
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