Given a series 1, 17, 98, 354 …… Find the nth term of this series.
The series basically represents the sum of the 4th power of first n natural numbers. First-term is the sum of 14. Second term is sum of two numbers i.e (14 + 24 = 17), third term i.e.(14 + 24 + 34 = 98) and so on.
Examples:
Input : 5 Output : 979 Input : 7 Output : 4676
Naive Approach :
A simple solution is to add the 4th powers of first n natural numbers. By using iteration we can easily find the nth term of the series.
Below is the implementation of the above approach :
C++
// CPP program to find n-th term of// series#include <iostream>using namespace std;// Function to find the nth term of seriesint sumOfSeries(int n){ // Loop to add 4th powers int ans = 0; for (int i = 1; i <= n; i++) ans += i * i * i * i; return ans;}// Driver codeint main(){ int n = 4; cout << sumOfSeries(n); return 0;} |
Java
// Java program to find n-th term of// seriesimport java.io.*;class GFG { // Function to find the nth term of series static int sumOfSeries(int n) { // Loop to add 4th powers int ans = 0; for (int i = 1; i <= n; i++) ans += i * i * i * i; return ans; } // Driver code public static void main(String args[]) { int n = 4; System.out.println(sumOfSeries(n)); }} |
Python3
# Python 3 program to find# n-th term of# series # Function to find the# nth term of seriesdef sumOfSeries(n) : # Loop to add 4th powers ans = 0 for i in range(1, n + 1) : ans = ans + i * i * i * i return ans # Driver coden = 4print(sumOfSeries(n)) |
C#
// C# program to find n-th term of// seriesusing System;class GFG { // Function to find the // nth term of series static int sumOfSeries(int n) { // Loop to add 4th powers int ans = 0; for (int i = 1; i <= n; i++) ans += i * i * i * i; return ans; } // Driver code public static void Main() { int n = 4; Console.WriteLine(sumOfSeries(n)); }}// This code is contributed by anuj_67 |
PHP
<?php// PHP program to find // n-th term of series// Function to find the // nth term of seriesfunction sumOfSeries( $n){ // Loop to add 4th powers $ans = 0; for ( $i = 1; $i <= $n; $i++) $ans += $i * $i * $i * $i; return $ans;}// Driver code$n = 4;echo sumOfSeries($n);// This code is contributed// by anuj_67?> |
Javascript
<script>// Javascript program to find n-th term of// series// Function to find the nth term of seriesfunction sumOfSeries( n){ // Loop to add 4th powers let ans = 0; for (let i = 1; i <= n; i++) ans += i * i * i * i; return ans;}// Driver Codelet n = 4;document.write(sumOfSeries(n));</script> |
Output :
354
Time Complexity : O(n).
Auxiliary Space: O(1)
Efficient approach :
The pattern in this series is nth term is equal to the sum of (n-1)th term and n4.
Examples:
n = 2 2nd term equals to sum of 1st term and 24 i.e 16 A2 = A1 + 16 = 1 + 16 = 17 Similarly, A3 = A2 + 34 = 17 + 81 = 98 and so on..
We get:
A(n) = A(n - 1) + n4
= A(n - 2) + n4 + (n-1)4
= A(n - 3) + n4 + (n-1)4 + (n-2)4
.
.
.
= A(1) + 16 + 81... + (n-1)4 + n4
A(n) = 1 + 16 + 81 +... + (n-1)4 + n4
= n(n + 1)(6n3 + 9n2 + n - 1) / 30
i.e A(n) is sum of 4th powers of First n natural numbers.
Below is the implementation of the above approach:
C++
// CPP program to find the n-th// term in series#include <bits/stdc++.h>using namespace std;// Function to find nth termint sumOfSeries(int n){ return n * (n + 1) * (6 * n * n * n + 9 * n * n + n - 1) / 30;}// Driver codeint main(){ int n = 4; cout << sumOfSeries(n); return 0;} |
Java
// Java program to find the n-th// term in seriesimport java.io.*;class Series { // Function to find nth term static int sumOfSeries(int n) { return n * (n + 1) * (6 * n * n * n + 9 * n * n + n - 1) / 30; } // Driver Code public static void main(String[] args) { int n = 4; System.out.println(sumOfSeries(n)); }} |
Python
# Python program to find the Nth # term in series # Function to print nth term # of series def sumOfSeries(n): return n * (n + 1) * (6 * n * n * n + 9 * n * n + n - 1)/ 30 # Driver code n = 4print sumOfSeries(n) |
C#
// C# program to find the n-th// term in seriesusing System;class Series { // Function to find nth term static int sumOfSeries(int n) { return n * (n + 1) * (6 * n * n * n + 9 * n * n + n - 1) / 30; } // Driver Code public static void Main() { int n = 4; Console.WriteLine(sumOfSeries(n)); }}// This code is contributed by anuj_67 |
PHP
<?php// PHP program to find the n-th// term in series// Function to find nth termfunction sumOfSeries( $n){ return $n * ($n + 1) * (6 * $n * $n * $n + 9 * $n * $n + $n - 1) / 30;} // Driver code $n = 4; echo sumOfSeries($n);// This code is contributed by anuj_67?> |
Javascript
<script> // Javascript program to find the n-th term in series // Function to find nth term function sumOfSeries(n) { return n * (n + 1) * (6 * n * n * n + 9 * n * n + n - 1) / 30; } let n = 4; document.write(sumOfSeries(n));// This code is contributed by divyeshrabadiya07.</script> |
Output:
354
Time Complexity : O(1).
Auxiliary Space: O(1)
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