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.