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

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++
Java
Python3
C#
PHP
Javascript
// 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 Codeint main(){    cout << sumofseries(3) << endl;    return 0;}
// 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));    }}
# 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 Programprint(sumofseries(3)) # This code is contributed# by Smitha Dinesh Semwal
// 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 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.?>
<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>
90
 
Complexity : O(N)
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)/2Sn = ((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++
Java
Python3
C#
PHP
Javascript
// 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 seriesint sumofseries(int n){    return (n * (n + 1) *           (n + 2) * (n + 3) / 4);} // Driver Codeint main(){    cout << sumofseries(3) << endl;    return 0;}
// 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));    }}
# Efficient CPP program to find sum of the# series 1.2.3 + 2.3.4 + 3.4.5 + ... # function to calculate sum of seriesdef sumofseries(n):     return int(n * (n + 1) * (n + 2) * (n + 3) / 4)  # Driver programprint(sumofseries(3))      # This code is contributed# by Smitha Dinesh Semwal
// 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// Efficient CPP program// to find sum of the// series 1.2.3 + 2.3.4// + 3.4.5 + ... // function to calculate// sum of seriesfunction sumofseries($n){    return ($n * ($n + 1) *           ($n + 2) * ($n + 3) / 4);}     // Driver Code    echo sumofseries(3); // This code is contributed by anuj_67.?>
<script>// Efficient Javascript program// to find sum of the// series 1.2.3 + 2.3.4// + 3.4.5 + ... // function to calculate// sum of seriesfunction sumofseries(n){    return (n * (n + 1) *        (n + 2) * (n + 3) / 4);}     // Driver Code    document.write(sumofseries(3)); // This code is contributed by gfgking </script>
90
 
Time Complexity : O(1) 
vt_m
code_hunt
gfgking
number-theory
series
series-sum
Mathematical
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series
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