Program to find Sum of the series 1*3 + 3*5 + ….
Given a series:
Sn = 1*3 + 3*5 + 5*7 + …
It is required to find the sum of first n terms of this series represented by Sn, where n is given taken input.
Examples:
Input : n = 2 Output : S<sub>n</sub> = 18 Explanation: The sum of first 2 terms of Series is 1*3 + 3*5 = 3 + 15 = 18 Input : n = 4 Output : S<sub>n</sub> = 116 Explanation: The sum of first 4 terms of Series is 1*3 + 3*5 + 5*7 + 7*9 = 3 + 15 + 35 + 63 = 116
Let, the n-th term be denoted by tn.
This problem can easily be solved by observing that the nth term can be founded by following method:
tn = (n-th term of (1, 3, 5, … ) )*(nth term of (3, 5, 7, ….))
Now, n-th term of series 1, 3, 5 is given by 2*n-1
and, the n-th term of series 3, 5, 7 is given by 2*n+1
Putting these two values in tn:
tn = (2*n-1)*(2*n+1) = 4*n*n-1
Now, the sum of first n terms will be given by :
Sn = ∑(4*n*n – 1)
=∑4*{n*n}-∑(1)
Now, it is known that the sum of first n terms of series n*n (1, 4, 9, …) is given by: n*(n+1)*(2*n+1)/6
And sum of n number of 1’s is n itself.
Now, putting values in Sn:
Sn = 4*n*(n+1)*(2*n+1)/6 – n
= n*(4*n*n + 6*n – 1)/3
Now, Sn value can be easily found by putting the desired value of n.
Below is the implementation of the above approach:
C++
// C++ program to find sum of first n terms #include <bits/stdc++.h> using namespace std; int calculateSum( int n) { // Sn = n*(4*n*n + 6*n - 1)/3 return (n * (4 * n * n + 6 * n - 1) / 3); } int main() { // number of terms to be included in the sum int n = 4; // find the Sn cout << "Sum = " << calculateSum(n); return 0; } |
Java
// Java program to find sum // of first n terms class GFG { static int calculateSum( int n) { // Sn = n*(4*n*n + 6*n - 1)/3 return (n * ( 4 * n * n + 6 * n - 1 ) / 3 ); } // Driver Code public static void main(String args[]) { // number of terms to be // included in the sum int n = 4 ; // find the Sn System.out.println( "Sum = " + calculateSum(n)); } } // This code is contributed by Bilal |
Python
# Python program to find sum # of first n terms def calculateSum(n): # Sn = n*(4*n*n + 6*n - 1)/3 return (n * ( 4 * n * n + 6 * n - 1 ) / 3 ); # Driver Code # number of terms to be # included in the sum n = 4 # find the Sn print ( "Sum =" ,calculateSum(n)) # This code is contributed by Bilal |
C#
// C# program to find sum // of first n terms using System; class GFG { static int calculateSum( int n) { // Sn = n*(4*n*n + 6*n - 1)/3 return (n * (4 * n * n + 6 * n - 1) / 3); } // Driver code static public void Main () { // number of terms to be // included in the sum int n = 4; // find the Sn Console.WriteLine( "Sum = " + calculateSum(n)); } } // This code is contributed // by mahadev |
PHP
<?php // PHP program to find sum // of first n terms function calculateSum( $n ) { // Sn = n*(4*n*n + 6*n - 1)/3 return ( $n * (4 * $n * $n + 6 * $n - 1) / 3); } // number of terms to be // included in the sum $n = 4; // 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 calculateSum( n) { // Sn = n*(4*n*n + 6*n - 1)/3 return (n * (4 * n * n + 6 * n - 1) / 3); } // Driver Code // number of terms to be // included in the sum let n = 4; // find the Sn document.write( "Sum = " + calculateSum(n)); // This code contributed by Princi Singh </script> |
Sum = 116
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
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