Find the sum of first n terms of the given series:
3, 20, 63, 144, .....
Examples:
Input : n = 2 Output : 23 Input : n =4 Output : 230
Approach:
First, we have to find the general term (Tn) of the given series.
series can we written in the following way also: (3 * 1^2), (5 * 2^2), (7 * 3^2), (9 * 4^2), .......up t n terms Tn = (General term of series 3, 5, 7, 9 ....) X (General term of series 1^2, 2^2, 3^2, 4^2 ....) Tn = (3 + (n-1) * 2) X ( n^2 ) Tn = 2*n^3 + n^2
We can write the sum of the series in the following ways:
Sn = 3 + 20 + 63 + 144 + ........up to the n terms[Tex]$$ Sn = 2 \times \sum_{n=1}^{n} n^{3} + \sum_{n=1}^{n} n^{2} $$[/Tex]Sn = 2 * (sum of n terms of n^3 ) + (sum of n terms of n^2)
Following are the formulas of sum of n terms of n^3 and n^2 :
Below is the implementation of the above approach:
C++
// C++ program to find the sum of n terms #include <bits/stdc++.h> using namespace std;
int calculateSum( int n)
{ return (2 * pow ((n * (n + 1) / 2), 2)) +
((n * (n + 1) * (2 * n + 1)) / 6);
} int main()
{ int n = 4;
cout << "Sum = " << calculateSum(n) << endl;
return 0;
} |
Java
// Java program to find the sum of n terms import java.io.*;
public class GFG
{ static int calculateSum( int n)
{
return ( int )(( 2 * Math.pow((n * (n + 1 ) / 2 ), 2 ))) +
((n * (n + 1 ) * ( 2 * n + 1 )) / 6 );
}
public static void main (String[] args) {
int n = 4 ;
System.out.println( "Sum = " + calculateSum(n));
}
} // This code is contributed by Raj |
Python3
# Python3 program to find the sum of n terms def calculateSum(n):
return (( 2 * (n * (n + 1 ) / 2 ) * * 2 ) +
((n * (n + 1 ) * ( 2 * n + 1 )) / 6 ))
#Driver code n = 4
print ( "Sum =" ,calculateSum(n))
# this code is contributed by Shashank_Sharma |
C#
// C# program to find the sum of n terms using System;
class GFG
{ static int calculateSum( int n)
{ return ( int )((2 * Math.Pow((n * (n + 1) / 2), 2))) +
((n * (n + 1) * (2 * n + 1)) / 6);
} // Driver Code public static void Main ()
{ int n = 4;
Console.WriteLine( "Sum = " + calculateSum(n));
} } // This code is contributed by anuj_67 |
PHP
<?php // PHP program to find the // sum of n terms function calculateSum( $n )
{ return (2 * pow(( $n * ( $n + 1) / 2), 2)) +
(( $n * ( $n + 1) * (2 * $n + 1)) / 6);
} // Driver Code $n = 4;
echo "Sum = " , calculateSum( $n );
// This code is contributed by ash264 ?> |
Javascript
<script> // javascript program to find the sum of n terms function calculateSum(n)
{ return parseInt(((2 * Math.pow((n * (n + 1) / 2), 2))) +
((n * (n + 1) * (2 * n + 1)) / 6));
} var n = 4;
document.write( "Sum = " + calculateSum(n));
// This code contributed by shikhasingrajput </script> |
Output:
Sum = 230
Time Complexity: O(1)
Auxiliary Space: O(1)