Sum of the first N terms of the series 5,12, 23, 38….
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++
// 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
// 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 3
# 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#
// 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 // 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 ?> |
Javascript
<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> |
Output:
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.
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