Program to find sum of series 1*2*3 + 2*3*4+ 3*4*5 + . . . + n*(n+1)*(n+2)
Given a positive integer n and the task is to find the sum of series 1*2*3 + 2*3*4 + 4*5*6 + . . .+ n*(n+1)*(n+2).
Examples:
Input : n = 10 Output : 4290 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + 5*6*7 + 6*7*8 + 7*8*9 + 8*9*10 + 9*10*11 + 10*11*12 = 6 + 24 + 60 + 120 + 210 + 336 + 504 + 720 + 990 + 1320 = 4290 Input : n = 7 Output : 1260
Method 1: In this case loop will run n times and calculate the sum.
C++
// Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) #include <bits/stdc++.h> using namespace std; // Function to calculate sum of series. int sumOfSeries(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum = sum + i * (i + 1) * (i + 2); return sum; } // Driver function int main() { int n = 10; cout << sumOfSeries(n); return 0; } |
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Java
// Java Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) public class GfG{ // Function to calculate sum of series. static int sumOfSeries(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum = sum + i * (i + 1) * (i + 2); return sum; } // Driver Code public static void main(String s[]) { int n = 10; System.out.println(sumOfSeries(n)); } } // This article is contributed by Gitanjali. |
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Python3
# Python program to find the # sum of series # 1*2*3 + 2*3*4 + . . . # + n*(n+1)*(n+1) # Function to calculate sum # of series. def sumOfSeries(n): sum = 0; i = 1; while i<=n: sum = sum + i * (i + 1) * ( i + 2) i = i + 1 return sum # Driver code n = 10print(sumOfSeries(n)) # This code is contributed by "Abhishek Sharma 44" |
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C#
// C# Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) using System; public class GfG { // Function to calculate sum of series. static int sumOfSeries(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum = sum + i * (i + 1) * (i + 2); return sum; } // Driver Code public static void Main() { int n = 10; Console.WriteLine(sumOfSeries(n)); } } // This article is contributed by vt_m. |
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PHP
<?php // PHP Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) // Function to calculate sum of series. function sumOfSeries($n) { $sum = 0; for ($i = 1; $i <= $n; $i++) $sum = $sum + $i * ($i + 1) * ($i + 2); return $sum; } // Driver Code $n = 10; echo sumOfSeries($n); // This code is contributed by vt_m. ?> |
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Output:
4290
Time Complexity: O(n)
Method 2: In this case we use formula to add sum of series.
Given series 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6 + . . . + n*(n+1)*(n+2)
sum of series = (n * (n+1) * (n+2) * (n+3)) / 4
Put n = 10 then
sum = (10 * (10+1) * (10+2) * (10+3)) / 4
= (10 * 11 * 12 * 13) / 4
= 4290
C++
// Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) #include <bits/stdc++.h> using namespace std; // Function to calculate sum of series. int sumOfSeries(int n) { return (n * (n + 1) * (n + 2) * (n + 3)) / 4; } // Driver function int main() { int n = 10; cout << sumOfSeries(n); return 0; } |
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Java
// Program to find the // sum of series // 1*2*3 + 2*3*4 + // . . . + n*(n+1)*(n+1) import java.io.*; class GFG { // Function to calculate // sum of series. static int sumOfSeries(int n) { return (n * (n + 1) * (n + 2) * (n + 3)) / 4; } // Driver function public static void main (String[] args) { int n = 10; System.out.println(sumOfSeries(n)); } } // This code is contributed by Nikita Tiwari. |
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Python3
# Python program to find the # sum of series # 1*2*3 + 2*3*4 + . . . # + n*(n+1)*(n+1) # Function to calculate sum # of series. def sumOfSeries(n): return (n * (n + 1) * (n + 2 ) * (n + 3)) / 4 #Driver code n = 10print(sumOfSeries(n)) # This code is contributed by "Abhishek Sharma 44" |
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C#
// Program to find the // sum of series // 1*2*3 + 2*3*4 + // . . . + n*(n+1)*(n+1) using System; class GFG { // Function to calculate // sum of series. static int sumOfSeries(int n) { return (n * (n + 1) * (n + 2) * (n + 3)) / 4; } // Driver function public static void Main () { int n = 10; Console.WriteLine(sumOfSeries(n)); } } // This code is contributed by vt_m. |
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PHP
<?php // PHP Program to find the sum of series // 1*2*3 + 2*3*4 + . . . + n*(n+1)*(n+1) // Function to calculate sum of series. function sumOfSeries($n) { return ($n * ($n + 1) * ($n + 2) * ($n + 3)) / 4; } // Driver Code $n = 10; echo sumOfSeries($n); // This code is contributed by vt_m. ?> |
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Output:
4290
Time Complexity: O(1)
How does this formula work?
We can prove working of this formula using
mathematical induction.
According to formula, sum of (k -1) terms is
((k - 1) * (k) * (k + 1) * (k + 2)) / 4
Sum of k terms
= sum of k-1 terms + value of k-th term
= ((k - 1) * (k) * (k + 1) * (k + 2)) / 4 +
k * (k + 1) * (k + 2)
Taking common term (k + 1) * (k + 2) out.
= (k + 1)*(k + 2) [k*(k-1)/4 + k]
= (k + 1)*(k + 2) * k * (k + 3)/4
= k * (k + 1) * (k + 2) * (k + 3)/4
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