Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + …… + (1+3+5+7+…+(2n-1))
Given a positive integer n. The problem is to find the sum of the given series 1 + (1+2) + (1+2+3) + (1+2+3+4) + …… + (1+2+3+4+…+n), where i-th term in the series is the sum of first i odd natural numbers.
Examples:
Input : n = 2 Output : 5 (1) + (1+3) = 5 Input : n = 5 Output : 55 (1) + (1+3) + (1+3+5) + (1+3+5+7) + (1+3+5+7+9) = 55
Naive Approach: Using two loops get the sum of each i-th term and then add those sum to the final sum.
C++
// C++ implementation to find the // sum of the given series #include <bits/stdc++.h> using namespace std; // functionn to find the // sum of the given series int sumOfTheSeries( int n) { int sum = 0; for ( int i = 1; i <= n; i++) { // first term of each i-th term int k = 1; for ( int j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } // Driver program int main() { int n = 5; cout << "Sum = " << sumOfTheSeries(n); return 0; } |
Java
// Java implementation to find // the sum of the given series import java.util.*; class GFG { // functionn to find the sum // of the given series static int sumOfTheSeries( int n) { int sum = 0 ; for ( int i = 1 ; i <= n; i++) { // first term of each // i-th term int k = 1 ; for ( int j = 1 ; j <= i; j++) { sum += k; // next term k += 2 ; } } // required sum return sum; } /* Driver program */ public static void main(String[] args) { int n = 5 ; System.out.println( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 implementation to find # the sum of the given series # functionn to find the sum # of the given series def sumOfTheSeries( n ): sum = 0 for i in range ( 1 , n + 1 ): # first term of each i-th term k = 1 for j in range ( 1 ,i + 1 ): sum + = k # next term k + = 2 # required sum return sum # Driver program n = 5 print ( "Sum =" , sumOfTheSeries(n)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# implementation to find // the sum of the given series using System; class GFG { // functionn to find the sum // of the given series static int sumOfTheSeries( int n) { int sum = 0; for ( int i = 1; i <= n; i++) { // first term of each // i-th term int k = 1; for ( int j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } /* Driver program */ public static void Main() { int n = 5; Console.Write( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by vt_m. |
php
<?php // php implementation to find the // sum of the given series // functionn to find the // sum of the given series function sumOfTheSeries( $n ) { $sum = 0; for ( $i = 1; $i <= $n ; $i ++) { // first term of each i-th term $k = 1; for ( $j = 1; $j <= $i ; $j ++) { $sum += $k ; // next term $k += 2; } } // required sum return $sum ; } // Driver program $n = 5; echo "Sum = " . sumOfTheSeries( $n ); // This code is contributed by Sam007 ?> |
Javascript
<script> // Javascript implementation to find the // sum of the given series // functionn to find the // sum of the given series function sumOfTheSeries(n) { let sum = 0; for (let i = 1; i <= n; i++) { // first term of each i-th term let k = 1; for (let j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } // Driver program let n = 5; document.write( "Sum = " + sumOfTheSeries(n)); // This code is contributed by gfgking </script> |
Output:
Sum = 55
Efficient Approach:
Let an be the n-th term of the given series.
an = (1 + 3 + 5 + 7 + (2n-1)) = sum of first n odd numbers = n2
Refer this post for the proof of above formula.
Now,
Refer this post for the proof of above formula.
C++
// C++ implementation to find the sum // of the given series #include <bits/stdc++.h> using namespace std; // functionn to find the sum // of the given series int sumOfTheSeries( int n) { // required sum return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driver program to test above int main() { int n = 5; cout << "Sum = " << sumOfTheSeries(n); return 0; } |
Java
// Java implementation to find // the sum of the given series import java.io.*; class GfG { // function to find the sum // of the given series static int sumOfTheSeries( int n) { // required sum return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ; } // Driver program to test above public static void main (String[] args) { int n = 5 ; System.out.println( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by Gitanjali. |
Python3
# Python3 implementation to find # the sum of the given series # functionn to find the sum # of the given series def sumOfTheSeries( n ): # required sum return int ((n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ) # Driver program to test above n = 5 print ( "Sum =" , sumOfTheSeries(n)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# implementation to find // the sum of the given series using System; class GfG { // function to find the sum // of the given series static int sumOfTheSeries( int n) { // required sum return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driver program to test above public static void Main() { int n = 5; Console.Write( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP implementation to find the sum // of the given series // functionn to find the sum // of the given series function sumOfTheSeries( $n ) { // required sum return ( $n * ( $n + 1) / 2) * (2 * $n + 1) / 3; } // Driver Code $n = 5; echo "Sum = " . sumOfTheSeries( $n ); // This code is contributed by Sam007 ?> |
Javascript
<script> // JavaScript program to find // the sum of the given series // function to find the sum // of the given series function sumOfTheSeries(n) { // required sum return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driver Code let n = 5; document.write( "Sum = " + sumOfTheSeries(n)); // This code is contributed by avijitmondal1998. </script> |
Output:
Sum = 55