Find the nth term of the series 0, 8, 64, 216, 512, . . .
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: 1000Input: 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 series long 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 code int main() { int n = 5; cout << term(n); return 0; } |
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Java
// Java implementation of the approach import 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)); } } |
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Python3
# Python3 implementation of the approach # Function to return the nth term of the given series def 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 code n = 5print(term(n)) |
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C#
// C# implementation of the approach using 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. |
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PHP
<?php // PHP implementation of the approach // Function to return the nth term of the given series function 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); ?> |
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Output:
512
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