Given an integer N, the task is to find the Nth term of the following series:
0, 8, 64, 216, 512, 1000, 1728, . . .
Examples:
Input: N = 6
Output: 1000
Input: N = 5
Output: 512
Approach:
- Given series 0, 8, 64, 216, 512, 1000, 1728, … can also be written as 0 * (02), 2 * (22), 4 * (42), 6 * (62), 8 * (82), 10 * (102), …
- Observe that 0, 2, 4, 6, 10, … is in AP and the nth term of this series can be found using the formula term = a1 + (n – 1) * d where a1 is the first term, n is the term position and d is the common difference.
- To get the term in the original series, term = term * (term2) i.e. term3.
- Finally print the term.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the nth term of the given serieslong term(int n){ // Common difference int d = 2; // First term int a1 = 0; // nth term int An = a1 + (n - 1) * d; // nth term of the given series An = pow(An, 3); return An;}// Driver codeint main(){ int n = 5; cout << term(n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;import java.lang.*;import java.io.*;public class GFG { // Function to return the nth term of the given series static int nthTerm(int n) { // Common difference and first term int d = 2, a1 = 0; // nth term int An = a1 + (n - 1) * d; // nth term of the given series return (int)Math.pow(An, 3); } // Driver code public static void main(String[] args) { int n = 5; System.out.println(nthTerm(n)); }} |
Python3
# Python3 implementation of the approach# Function to return the nth term of the given seriesdef term(n): # Common difference d = 2 # First term a1 = 0 # nth term An = a1 +(n-1)*d # nth term of the given series An = An**3 return An; # Driver coden = 5print(term(n)) |
C#
// C# implementation of the approachusing System;public class GFG { // Function to return the nth term of the given series static int nthTerm(int n) { // Common difference and first term int d = 2, a1 = 0; // nth term int An = a1 + (n - 1) * d; // nth term of the given series return (int)Math.Pow(An, 3); } // Driver code public static void Main() { int n = 5; Console. WriteLine(nthTerm(n)); }}// This code is contributed by Mutual singh. |
PHP
<?php// PHP implementation of the approach// Function to return the nth term of the given seriesfunction term($n) { // Common difference $d = 2; // First term $a1 = 0; // nth term $An=$a1+($n-1)*$d; // nth term of the given series return pow($An, 3); } // Driver code $n = 5;echo term($n); ?> |
Javascript
<script>// javascript implementation of the approach// Function to return the nth term of the given seriesfunction term(n) { // Common difference let d = 2; // First term let a1 = 0; // nth term An=a1+(n-1)*d; // nth term of the given series return Math.pow(An, 3); } // Driver code let n = 5;document.write( term(n)); // This code is contributed by sravan kumar</script> |
Output:
512
Time Complexity: O(1)
Auxiliary Space: O(1)
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