Given a number N, the task is to find the sum of the first N terms of the below series:
Sn = 2 + 6 + 12 + 20 + 30 … upto n terms
Examples:
Input: N = 2 Output: 8 Explanation: 2 + 6 = 8 Input: N = 4 Output: 40 Explanation: 2 + 6+ 12 + 20 = 40
Approach: Let, the nth term be denoted by Sn.
This problem can easily be solved by splitting each term as follows :
Sn = 2 + 6 + 12 + 20 + 30...... Sn = (1+1^2) + (2+2^2) + (3+3^2) + (4+4^2) +...... Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms)
We observed that Sn can break down into summation of two series.
Hence, the sum of the first n terms is given as follows:
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms) Sn = n*(n + 1)/2 + n*(n + 1)*(2*n + 1)/6
Below is the implementation of the above approach:
C++
// C++ program to find sum of first n terms#include <bits/stdc++.h>using namespace std;// Function to calculate the sumint calculateSum(int n){ return n * (n + 1) / 2 + n * (n + 1) * (2 * n + 1) / 6;}// Driver codeint main(){ // number of terms to be // included in the sum int n = 3; // find the Sn cout << "Sum = " << calculateSum(n); return 0;} |
Java
// Java program to find sum of first n terms import java.util.*;import java.lang.*;import java.io.*;class GFG{ // Function to calculate the sum static int calculateSum(int n) { return n * (n + 1) / 2 + n * (n + 1) * (2 * n + 1) / 6; } // Driver code public static void main(String args[]) { // number of terms to be // included in the sum int n = 3; // find the Sn System.out.print("Sum = " + calculateSum(n)); }} |
Python3
# Python program to find sum of# first n terms# Function to calculate the sumdef calculateSum(n): return (n * (n + 1) // 2 + n * (n + 1) * (2 * n + 1) // 6)# Driver code# number of terms to be# included in the sumn = 3# find the Sumprint("Sum = ", calculateSum(n))# This code is contributed by# Sanjit_Prasad |
C#
// C# program to find sum of // first n terms using System;class GFG{ // Function to calculate the sum static int calculateSum(int n) { return n * (n + 1) / 2 + n * (n + 1) * (2 * n + 1) / 6; } // Driver code public static void Main() { // number of terms to be // included in the sum int n = 3; // find the Sn Console.WriteLine("Sum = " + calculateSum(n)); }} // This code is contributed by inder_verma |
PHP
<?php// PHP program to find sum// of first n terms// Function to calculate the sumfunction calculateSum($n){ return $n * ($n + 1) / 2 + $n * ($n + 1) * (2 * $n + 1) / 6;}// Driver code// number of terms to be// included in the sum$n = 3;// find the Snecho "Sum = " , calculateSum($n);// This code is contributed // by inder_verma?> |
Javascript
<script>// Javascript program to find sum of first n terms // Function to calculate the sum function calculateSum(n) { return n * (n + 1) / 2 + n * (n + 1) * (2 * n + 1) / 6; } // Driver code // number of terms to be // included in the sum let n = 3; // find the Sn document.write("Sum = " + calculateSum(n)); // This code is contributed by Mayank Tyagi</script> |
Output:
Sum = 20
Time Complexity: O(1), we are using only constant-time operations.
Auxiliary Space: O(1)
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