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Sum of the series 1.2.3 + 2.3.4 + … + n(n+1)(n+2)

  • Difficulty Level : Hard
  • Last Updated : 20 Apr, 2021

Find the sum up to n terms of the series: 1.2.3 + 2.3.4 + … + n(n+1)(n+2). In this 1.2.3 represent the first term and 2.3.4 represent the second term .
Examples : 
 

Input : 2
Output : 30
1.2.3 + 2.3.4 = 6 + 24 = 30

Input : 3
Output : 90

 

Simple Approach We run a loop for i = 1 to n, and find the sum of (i)*(i+1)*(i+2). 
And at the end display the sum .
 

C++




// CPP program to find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
#include <bits/stdc++.h>
using namespace std;
 
int sumofseries(int n)
{
    int res = 0;
    for (int i = 1; i <= n; i++)
        res += (i) * (i + 1) * (i + 2);   
    return res;
}
 
// Driver Code
int main()
{
    cout << sumofseries(3) << endl;
    return 0;
}

Java




// Java program to find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
import java.io.*;
import java.math.*;
 
class GFG
{
 
    static int sumofseries(int n)
    {
    int res = 0;
    for (int i = 1; i <= n; i++)
        res += (i) * (i + 1) * (i + 2);
    return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(sumofseries(3));
    }
}

Python3




# Python 3 program to find sum of the series
# 1.2.3 + 2.3.4 + 3.4.5 + ...
 
def sumofseries(n):
 
    res = 0
    for i in range(1, n+1):
        res += (i) * (i + 1) * (i + 2)
    return res
 
# Driver Program
print(sumofseries(3))
 
# This code is contributed
# by Smitha Dinesh Semwal

C#




// Java program to find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
using System;
 
class GFG
{
 
    static int sumofseries(int n)
    {
        int res = 0;
        for (int i = 1; i <= n; i++)
            res += (i) * (i + 1) * (i + 2);
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
        Console.WriteLine(sumofseries(3));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find
// sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
 
function sumofseries($n)
{
    $res = 0;
    for ($i = 1; $i <= $n; $i++)
        $res += ($i) * ($i + 1) *
                       ($i + 2);
    return $res;
}
 
    // Driver Code
    echo sumofseries(3);
 
//This code is contributed by anuj_67.
?>

Javascript




<script>
// JavaScript program to find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
 
    function sumofseries(n)
    {
    let res = 0;
    for (let i = 1; i <= n; i++)
        res += (i) * (i + 1) * (i + 2);
    return res;
    }
 
// Driver Code
 
    document.write(sumofseries(3));
 
// This code is contributed by code_hunt.
</script>
Output : 
90

 

Complexity : O(N)

Efficient Approach

Using Efficient Approach we know that we have to find = summation of( (n)*(n+1)*(n+2) )
 

Sn = summation[ (n)*(n+1)*(n+2) ] 
Sn = summation [n3 + 2*n2 + n2 + 2*n]
We know sum of cubes of natural numbers is (n*(n+1))/2)2, sum of squares of natural numbers is n * (n + 1) * (2n + 1) / 6 and sum of first n natural numbers is n(n+1)/2
Sn = ((n*(n+1))/2)2 + 3((n)*(n+1)*(2*n+1)/6) + 2*((n)*(n+1)/2) 
So by evaluating the above we get, 
Sn = (n*(n+1)*(n+2)*(n+3)/4) 
Hence it has a O(1) complexity.

 

C++




// Efficient CPP program to
// find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ...
#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 Code
int main()
{
    cout << sumofseries(3) << endl;
    return 0;
}

Java




// Efficient Java program to
// find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ..
import java.io.*;
import java.math.*;
 
class GFG
{
    static int sumofseries(int n)
    {
    return (n * (n + 1) *
           (n + 2) * (n + 3) / 4);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(sumofseries(3));
    }
}

Python3




# Efficient CPP program to find sum of the
# series 1.2.3 + 2.3.4 + 3.4.5 + ...
 
# function to calculate sum of series
def sumofseries(n):
 
    return int(n * (n + 1) * (n + 2) * (n + 3) / 4)
 
 
# Driver program
print(sumofseries(3))
     
 
# This code is contributed
# by Smitha Dinesh Semwal

C#




// Efficient C# program to
// find sum of the series
// 1.2.3 + 2.3.4 + 3.4.5 + ..
using System;
 
class GFG
{
    static int sumofseries(int n)
    {
    return (n * (n + 1) *
           (n + 2) * (n + 3) / 4);
    }
 
    // Driver Code
    public static void Main()
    {
        Console.WriteLine(sumofseries(3));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// Efficient CPP program
// to find sum of the
// series 1.2.3 + 2.3.4
// + 3.4.5 + ...
 
// function to calculate
// sum of series
function sumofseries($n)
{
    return ($n * ($n + 1) *
           ($n + 2) * ($n + 3) / 4);
}
 
    // Driver Code
    echo sumofseries(3);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
// Efficient Javascript program
// to find sum of the
// series 1.2.3 + 2.3.4
// + 3.4.5 + ...
 
// function to calculate
// sum of series
function sumofseries(n)
{
    return (n * (n + 1) *
        (n + 2) * (n + 3) / 4);
}
 
    // Driver Code
    document.write(sumofseries(3));
 
// This code is contributed by gfgking
 
</script>
Output : 
90

 

Time Complexity : O(1)
 


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