Sum of the series 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13.. till N-th term
Given a series of numbers 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13… The task is to find the sum of all the numbers in series till N-th number.
Examples:
Input: N = 4
Output: 10
1 + 2 + 4 + 3 = 10
Input: N = 10
Output: 55
Approach: The series is basically 20 odd numbers, 21 even numbers, 22 even numbers…. The sum of first N odd numbers is N * N and sum of first N even numbers is (N * (N+1)). Calculate the summation for 2i odd or even numbers and keep adding them to the sum.
Iterate for every power of 2, till the number of iterations exceeds N, and keep adding the respective summation of odd or even numbers according to the parity. For every segment the sum of the segment will be, (current sum of X odd/even numbers – previous sum of Y odd/even numbers), where X is the total sum of odd/even numbers till this segment and Y is the summation of odd/even numbers till the previous when odd/even numbers occurred.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the// sum of first N odd numbersint sumodd(int n){ return (n * n);}// Function to find the// sum of first N even numbersint sumeven(int n){ return (n * (n + 1));}// Function to overall// find the sum of seriesint findSum(int num){ // Initiall odd numbers int sumo = 0; // Initial even numbers int sume = 0; // First power of 2 int x = 1; // Check for parity // for odd/even int cur = 0; // Counts the sum int ans = 0; while (num > 0) { // Get the minimum // out of remaining num // or power of 2 int inc = min(x, num); // Decrease that much numbers // from num num -= inc; // If the segment has odd numbers if (cur == 0) { // Summate the odd numbers // By exclusion ans = ans + sumodd(sumo + inc) - sumodd(sumo); // Increase number of odd numbers sumo += inc; } // If the segment has even numbers else { // Summate the even numbers // By exclusion ans = ans + sumeven(sume + inc) - sumeven(sume); // Increase number of even numbers sume += inc; } // Next set of numbers x *= 2; // Change parity for odd/even cur ^= 1; } return ans;}// Driver codeint main(){ int n = 4; cout << findSum(n); return 0;} |
Java
// Java program to implement// the above approachclass GFG{ // Function to find the // sum of first N odd numbers static int sumodd(int n) { return (n * n); } // Function to find the // sum of first N even numbers static int sumeven(int n) { return (n * (n + 1)); } // Function to overall // find the sum of series static int findSum(int num) { // Initiall odd numbers int sumo = 0; // Initial even numbers int sume = 0; // First power of 2 int x = 1; // Check for parity // for odd/even int cur = 0; // Counts the sum int ans = 0; while (num > 0) { // Get the minimum // out of remaining num // or power of 2 int inc = Math.min(x, num); // Decrease that much numbers // from num num -= inc; // If the segment has odd numbers if (cur == 0) { // Summate the odd numbers // By exclusion ans = ans + sumodd(sumo + inc) - sumodd(sumo); // Increase number of odd numbers sumo += inc; } // If the segment has even numbers else { // Summate the even numbers // By exclusion ans = ans + sumeven(sume + inc) - sumeven(sume); // Increase number of even numbers sume += inc; } // Next set of numbers x *= 2; // Change parity for odd/even cur ^= 1; } return ans; } // Driver code public static void main(String[] args) { int n = 4; System.out.println(findSum(n)); }}// This code contributed by Rajput-Ji |
Python
# Python3 program to implement# the above approach# Function to find the# sum of first N odd numbersdef sumodd(n): return (n * n)# Function to find the# sum of first N even numbersdef sumeven(n): return (n * (n + 1))# Function to overall# find the sum of seriesdef findSum(num): # Initiall odd numbers sumo = 0 # Initial even numbers sume = 0 # First power of 2 x = 1 # Check for parity # for odd/even cur = 0 # Counts the sum ans = 0 while (num > 0): # Get the minimum # out of remaining num # or power of 2 inc = min(x, num) # Decrease that much numbers # from num num -= inc # If the segment has odd numbers if (cur == 0): # Summate the odd numbers # By exclusion ans = ans + sumodd(sumo + inc) - sumodd(sumo) # Increase number of odd numbers sumo += inc # If the segment has even numbers else: # Summate the even numbers # By exclusion ans = ans + sumeven(sume + inc) - sumeven(sume) # Increase number of even numbers sume += inc # Next set of numbers x *= 2 # Change parity for odd/even cur ^= 1 return ans# Driver coden = 4print(findSum(n))# This code is contributed by mohit kumar |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to find the // sum of first N odd numbers static int sumodd(int n) { return (n * n); } // Function to find the // sum of first N even numbers static int sumeven(int n) { return (n * (n + 1)); } // Function to overall // find the sum of series static int findSum(int num) { // Initiall odd numbers int sumo = 0; // Initial even numbers int sume = 0; // First power of 2 int x = 1; // Check for parity // for odd/even int cur = 0; // Counts the sum int ans = 0; while (num > 0) { // Get the minimum // out of remaining num // or power of 2 int inc = Math.Min(x, num); // Decrease that much numbers // from num num -= inc; // If the segment has odd numbers if (cur == 0) { // Summate the odd numbers // By exclusion ans = ans + sumodd(sumo + inc) - sumodd(sumo); // Increase number of odd numbers sumo += inc; } // If the segment has even numbers else { // Summate the even numbers // By exclusion ans = ans + sumeven(sume + inc) - sumeven(sume); // Increase number of even numbers sume += inc; } // Next set of numbers x *= 2; // Change parity for odd/even cur ^= 1; } return ans; } // Driver code public static void Main(String[] args) { int n = 4; Console.WriteLine(findSum(n)); }}// This code has been contributed by 29AjayKumar |
PHP
<?php// PHP program to implement// the above approach// Function to find the// sum of first N odd numbersfunction sumodd($n){ return ($n * $n);}// Function to find the// sum of first N even numbersfunction sumeven($n){ return ($n * ($n + 1));}// Function to overall// find the sum of seriesfunction findSum( $num){ // Initiall odd numbers $sumo = 0; // Initial even numbers $sume = 0; // First power of 2 $x = 1; // Check for parity // for odd/even $cur = 0; // Counts the sum $ans = 0; while ($num > 0) { // Get the minimum // out of remaining num // or power of 2 $inc = min($x, $num); // Decrease that much numbers // from num $num -= $inc; // If the segment has odd numbers if ($cur == 0) { // Summate the odd numbers // By exclusion $ans = $ans + sumodd($sumo + $inc) - sumodd($sumo); // Increase number of odd numbers $sumo += $inc; } // If the segment has even numbers else { // Summate the even numbers // By exclusion $ans = $ans + sumeven($sume + $inc) - sumeven($sume); // Increase number of even numbers $sume += $inc; } // Next set of numbers $x *= 2; // Change parity for odd/even $cur ^= 1; } return $ans;}// Driver code$n = 4;echo findSum($n);// This code contributed by princiraj1992?> |
Javascript
<script>// javascript program to implement// the above approach// Function to find the// sum of first N odd numbersfunction sumodd( n){ return (n * n);}// Function to find the// sum of first N even numbersfunction sumeven( n){ return (n * (n + 1));}// Function to overall// find the sum of seriesfunction findSum( num){ // Initiall odd numbers let sumo = 0; // Initial even numbers let sume = 0; // First power of 2 let x = 1; // Check for parity // for odd/even let cur = 0; // Counts the sum let ans = 0; while (num > 0) { // Get the minimum // out of remaining num // or power of 2 let inc = Math.min(x, num); // Decrease that much numbers // from num num -= inc; // If the segment has odd numbers if (cur == 0) { // Summate the odd numbers // By exclusion ans = ans + sumodd(sumo + inc) - sumodd(sumo); // Increase number of odd numbers sumo += inc; } // If the segment has even numbers else { // Summate the even numbers // By exclusion ans = ans + sumeven(sume + inc) - sumeven(sume); // Increase number of even numbers sume += inc; } // Next set of numbers x *= 2; // Change parity for odd/even cur ^= 1; } return ans;}// Driver code let n = 4; document.write( findSum(n)); // This code is contributed by todaysgaurav</script> |
Output:
10
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