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n-th term of series 1, 17, 98, 354……

  • Last Updated : 27 May, 2021

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 series
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
int main()
{
    int n = 4;
    cout << sumOfSeries(n);
    return 0;
}

Java




// Java program to find n-th term of
// series
import 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 series
def 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 code
n = 4
print(sumOfSeries(n))

C#




// C# program to find n-th term of
// series
using 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 series
function 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 series
function 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 Code
 
let n = 4;
document.write(sumOfSeries(n));
 
 
</script>
Output : 
354

 

Time Complexity : O(n).



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 term
int sumOfSeries(int n)
{
    return n * (n + 1) * (6 * n * n * n
                 + 9 * n * n + n - 1) / 30;
}
 
// Driver code
int main()
{
    int n = 4;
    cout << sumOfSeries(n);
    return 0;
}

Java




// Java program to find the n-th
// term in series
import 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 = 4
print sumOfSeries(n)

C#




// C# program to find the n-th
// term in series
using 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 term
function 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).
 

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