Given a number N, the task is to find the sum of the below series till N terms.
Examples:
Input: N = 2
Output: 3
1 + 2 = 3
Input: N = 5
Output: 701
1 + 2 + 9 + 64 + 625 = 701
Approach: From the given series, find the formula for Nth term:
1st term = 1 = 11-1 2nd term = 2 = 22-1 3rd term = 9 = 33-1 4th term = 64 = 44-1 . . Nth term = NN - 1
Therefore:
Nth term of the series
Then iterate over numbers in the range [1, N] to find all the terms using the above formula and compute their sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the sum of series void printSeriesSum( int N)
{ long long sum = 0;
for ( int i = 1; i <= N; i++) {
// Generate the ith term and
// add it to the sum
sum += pow (i, i - 1);
}
// Print the sum
cout << sum << endl;
} // Driver Code int main()
{ int N = 5;
printSeriesSum(N);
return 0;
} |
Java
// Java program for the above approach class GFG{
// Function to find the sum of series static void printSeriesSum( int N)
{ long sum = 0 ;
for ( int i = 1 ; i <= N; i++) {
// Generate the ith term and
// add it to the sum
sum += Math.pow(i, i - 1 );
}
// Print the sum
System.out.print(sum + "\n" );
} // Driver Code public static void main(String[] args)
{ int N = 5 ;
printSeriesSum(N);
} } // This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach # Function to find the summ of series def printSeriessumm(N):
summ = 0
for i in range ( 1 ,N + 1 ):
# Generate the ith term and
# add it to the summ
summ + = pow (i, i - 1 )
# Print the summ
print (summ)
# Driver Code N = 5
printSeriessumm(N) # This code is contributed by shubhamsingh10 |
C#
// C# program for the above approach using System;
class GFG{
// Function to find the sum of series static void printSeriesSum( int N)
{ double sum = 0;
for ( int i = 1; i <= N; i++) {
// Generate the ith term and
// add it to the sum
sum += Math.Pow(i, i - 1);
}
// Print the sum
Console.Write(sum + "\n" );
} // Driver Code public static void Main(String[] args)
{ int N = 5;
printSeriesSum(N);
} } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // javascript program for the above approach // Function to find the sum of series function printSeriesSum( N)
{ let sum = 0;
for (let i = 1; i <= N; i++) {
// Generate the ith term and
// add it to the sum
sum += Math.pow(i, i - 1);
}
// Print the sum
document.write(sum);
} // Driver Code let N = 5;
printSeriesSum(N);
// This code is contributed by todaysgaurav </script> |
Output:
701
Time Complexity: O(N * log N)
Auxiliary Space: O(1), since no extra space has been taken.