Given a number N, the task is to find the sum of first N terms of the below series:
Sn = 2 + 10 + 30 + 68 + … upto n terms
Examples:
Input: N = 2 Output: 12 2 + 10 = 12 Input: N = 4 Output: 40 2 + 10 + 30 + 68 = 110
Approach: Let, the nth term be denoted by tn.
This problem can easily be solved by splitting each term as follows :
Sn = 2 + 10 + 30 + 68 + ...... Sn = (1+1^3) + (2+2^3) + (3+3^3) + (4+4^3) +...... Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)
We observed that Sn can broken down into summation of two series.
Hence, the sum of first n terms is given as follows:
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms) Sn = n*(n + 1)/2 + (n*(n + 1)/2)^2
Below is the implementation of 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 + pow((n * (n + 1) / 2), 2);}// Driver codeint main(){ // number of terms to be // included in the sum int n = 3; // find the Sum cout << "Sum = " << calculateSum(n); return 0;} |
Java
// Java program to find sum of first n terms public class GFG { // Function to calculate the sum static int calculateSum(int n) { return n * (n + 1) / 2 + (int)Math.pow((n * (n + 1) / 2), 2); } // Driver code public static void main(String args[]) { // number of terms to be // included in the sum int n = 3; // find the Sum System.out.println("Sum = "+ calculateSum(n)); } // This Code is contributed by ANKITRAI1} |
Python3
# Python program to find sum # of first n terms# Function to calculate the sumdef calculateSum(n): return (n * (n + 1) // 2 + pow((n * (n + 1) // 2), 2))# 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 termsusing System;class gfg{ // Function to calculate the sum public void calculateSum(int n) { double r = (n * (n + 1) / 2 + Math.Pow((n * (n + 1) / 2), 2)); Console.WriteLine("Sum = " + r); } // Driver code public static int Main() { gfg g = new gfg(); // number of terms to be // included in the sum int n = 3; // find the Sum g.calculateSum(n); Console.Read(); return 0; }} |
PHP
<?php// PHP program to find sum// of first n terms// Function to calculate the sumfunction calculateSum($n){ return $n * ($n + 1) / 2 + pow(($n * ($n + 1) / 2), 2);}// Driver code// number of terms to be// included in the sum$n = 3;// find the Sumecho "Sum = " , calculateSum($n);// This code is contributed // by anuj_67?> |
Javascript
<script>// Javascript program to find sum of first n terms // Function to calculate the sum function calculateSum(n) { return n * (n + 1) / 2 + Math.pow((n * (n + 1) / 2), 2); } // Driver code // number of terms to be // included in the sum let n = 3; // find the Sum document.write("Sum = " + calculateSum(n)); // This code is contributed by Mayank Tyagi</script> |
Output:
Sum = 42
Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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