Given a number N, the task is to find the sum of first N terms of the below series:
Sn = 5 + 12 + 23 + 38 + … upto n terms
Examples:
Input: N = 2 Output: 17 5 + 12 = 17 Input: N = 4 Output: 80 5 + 12 + 23 + 38 = 78
Approach: Let, the nth term be denoted by tn.
This problem can easily with the help of a general formula for these type of series,
The series given above is a quadratic series. They are special because the difference of consecutive terms of this series will be in arithmetic progression.
There general formula is given by:
General Formula = a*(n^2) + b*n + c
Now, by putting first 3 terms of series in general formula we can get values of a, b and c.
Sn = 5 + 12 + 30 + 68 + ...... tn = 2 * (n^2) + n + 2 Sn = 2 * (n * (n+1) * (2 * n+1)/6) + n * (n+1)/2 + 2 * (n)
Below is the implementation of above approach:
// 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 2 * (n * (n + 1) * (2 * n + 1) / 6)
+ n * (n + 1) / 2 + 2 * (n);
} // Driver code int main()
{ // number of terms to be included in sum
int n = 3;
// find the Sn
cout << "Sum = " << calculateSum(n);
return 0;
} |
// Java program to find sum of first n terms import java.io.*;
class GFG {
// Function to calculate the sum static int calculateSum( int n)
{ return 2 * (n * (n + 1 ) * ( 2 * n + 1 ) / 6 )
+ n * (n + 1 ) / 2 + 2 * (n);
} // Driver code public static void main (String[] args) {
// number of terms to be included in sum
int n = 3 ;
// find the Sn
System.out.print( "Sum = " + calculateSum(n));
}
} // This code is contributed // by anuj_67.. |
# Python program to find # sum of first n terms # Function to calculate the sum def calculateSum(n) :
return ( 2 * (n * (n + 1 ) *
( 2 * n + 1 ) / / 6 ) + n *
(n + 1 ) / / 2 + 2 * (n))
# Driver code if __name__ = = "__main__" :
# number of terms to be
# included in sum
n = 3
# find the Sn
print ( "Sum =" ,calculateSum(n))
# This code is contributed by ANKITRAI1 |
// C# program to find sum // of first n terms using System;
class GFG
{ // Function to calculate the sum static int calculateSum( int n)
{ return 2 * (n * (n + 1) * (2 * n + 1) / 6) +
n * (n + 1) / 2 + 2 * (n);
} // Driver code public static void Main ()
{ // number of terms to be
// included in sum
int n = 3;
// find the Sn
Console.WriteLine( "Sum = " + calculateSum(n));
} } // This code is contributed // by Shashank |
<?php // PHP program to find sum // of first n terms // Function to calculate the sum function calculateSum( $n )
{ return 2 * ( $n * ( $n + 1) *
(2 * $n + 1) / 6) +
$n * ( $n + 1) /
2 + 2 * ( $n );
} // Driver code // number of terms to // be included in sum $n = 3;
// find the Sn echo "Sum = " . calculateSum( $n );
// This code is contributed // by ChitraNayal ?> |
<script> // Javascript program to find sum of first n terms // Function to calculate the sum function calculateSum(n)
{ return 2 * (n * (n + 1) * (2 * n + 1) / 6)
+ n * (n + 1) / 2 + 2 * (n);
} // Driver code // number of terms to be included in sum
let n = 3;
// find the Sn
document.write( "Sum = " + calculateSum(n));
// This code is contributed by Mayank Tyagi </script> |
Sum = 40
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.