n-ésimo término de la serie 1, 17, 98, 354……

Dada una serie 1, 17, 98, 354 …… Encuentra el término n de esta serie.
La serie básicamente representa la suma de la cuarta potencia de los primeros n números naturales. El primer término es la suma de 1 ⁴ . El segundo término es la suma de dos números, es decir (1 ⁴ + 2 ⁴ = 17), el tercer término es decir (1 ⁴ + 2 ⁴ + 3 ⁴ = 98) y así sucesivamente.

Ejemplos:  

Input : 5
Output : 979

Input : 7
Output : 4676 

Enfoque ingenuo: 
una solución simple es sumar las cuartas potencias de los primeros n números naturales. Al usar la iteración, podemos encontrar fácilmente el término n de la serie.
A continuación se muestra la implementación del enfoque anterior: 

// 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 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));
    }
}

# 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# 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 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
?>

<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>

354

Complejidad temporal: O(n).

Enfoque eficiente: 
el patrón en esta serie es que el término n es igual a la suma del término (n-1) y n ⁴ . 

Ejemplos: 

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..

Obtenemos: 

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.

A continuación se muestra la implementación del enfoque anterior: 

// 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 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 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# 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 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
?>

<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>

354

Complejidad temporal: O(1).