Sum of series 1*1*2! + 2*2*3! + ……..+ n*n*(n+1)!
Given n, we need to find sum of 1*1*2! + 2*2*3! + ……..+ n*n*(n+1)!
Examples:
Input: 1
Output: 2
Input: 3
Output: 242
We may assume that overflow does not happen.
A simple solution is to compute terms one by one and add to result.
An efficient solution is based on direct formula 2 + (n*n + n – 2) * (n + 1)!
The working of formula is based on this post.
C++
// CPP program to find sum of the series. #include <bits/stdc++.h> using namespace std; int factorial( int n) { int res = 1; for ( int i = 2; i <= n; i++) res = res * i; return res; } // Function to calculate required series int calculateSeries( int n) { return 2 + (n * n + n - 2) * factorial(n + 1); } // Drivers code int main() { int n = 3; cout << calculateSeries(n); return 0; } |
Java
// java program to find sum of the series. import java.io.*; class GFG { static int factorial( int n) { int res = 1 ; for ( int i = 2 ; i <= n; i++) res = res * i; return res; } // Function to calculate required series static int calculateSeries( int n) { return 2 + (n * n + n - 2 ) * factorial(n + 1 ); } // Drivers code public static void main (String[] args) { int n = 3 ; System.out.println(calculateSeries(n)); } } // This code is contributed by anuj_67. |
Python3
# Python program to find sum of # the series. import math def factorial(n): res = 1 i = 2 for i in (n + 1 ): res = res * i return res # Function to calculate required # series def calculateSeries(n): return ( 2 + (n * n + n - 2 ) * math.factorial(n + 1 )) # Driver code n = 3 print (calculateSeries(n)) # This code is contributed by # Prateek bajaj |
C#
// C# program to find sum of the series. using System; class GFG { static int factorial( int n) { int res = 1; for ( int i = 2; i <= n; i++) res = res * i; return res; } // Function to calculate required series static int calculateSeries( int n) { return 2 + (n * n + n - 2) * factorial(n + 1); } // Driver code public static void Main () { int n = 3; Console.WriteLine(calculateSeries(n)); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP program to find sum of the series. function factorial( $n ) { $res = 1; for ( $i = 2; $i <= $n ; $i ++) $res = $res * $i ; return $res ; } // Function to calculate required series function calculateSeries( $n ) { return 2 + ( $n * $n + $n - 2) * factorial( $n + 1); } // Driver code $n = 3; echo calculateSeries( $n ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // java script program to find sum of the series. function factorial( n) { let res = 1; for (let i = 2; i <= n; i++) res = res * i; return res; } // Function to calculate required series function calculateSeries( n) { return 2 + (n * n + n - 2) * factorial(n + 1); } // Drivers code let n = 3; document.write(calculateSeries(n)); // This code is contributed by mohan pavan </script> |
Output:
242
Time Complexity : O(n)
Auxiliary Space: O(1)
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