Sum of all i such that ( 2 ^ i + 1 ) % 3 = 0 where i is in range [ 1 , n ] | C ++ implementation of the approach ; Function to return the required sum ; Total odd numbers from 1 to n ; Sum of first n odd numbers ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; int sumN ( int n ) { n = ( n + 1 ) / 2 ; return ( n * n ) ; } int main ( ) { int n = 3 ; cout << sumN ( n ) ; return 0 ; }