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 sum int calculateSum( int n) { return n * (n + 1) / 2 + pow ((n * (n + 1) / 2), 2); } // Driver code int 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 sum def 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 Sum print ( "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 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 sum function 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 Sum echo "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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