Find the sum of all the terms in the nth row of the series given below.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 .......................... ............................ (so on)
Examples:
Input : n = 2 Output : 18 terms in 2nd row and their sum sum = (3 + 4 + 5 + 6) = 18 Input : n = 4 Output : 132
Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum.
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula:
Sum(n) = n * (2 * n2 + 1)
The proof for the formula is given below:
Prerequisite:
- Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as:
Sum = (n * [2*a + (n-1)*d]) / 2
- Sum of 1st n natural numbers is given as:
Sum = (n * (n + 1)) / 2
Proof:
Let the number of terms from the beginning till the end of the nth row be p. Here p = 2 + 4 + 6 + .....n terms For the given AP series, a = 2, d = 2. Using the above formula for the sum of n terms of the AP series, we get, p = n * (n + 1) Similarly, let the number of terms from the beginning till the end of the (n-1)th row be q. Here q = 2 + 4 + 6 + .....n-1 terms For the given AP series, a = 2, d = 2. Using the above formula for the sum of n-1 terms of the AP series, we get, q = n * (n - 1) Now, Sum of all the terms in the nth row = sum of 1st p natural numbers - sum of 1st q natural numbers = (p * (p + 1)) / 2 - (q * (q + 1)) / 2 Substituting the values of p and q and then solving the equation, we will get, Sum of all the terms in the nth row = n * (2 * n2 + 1)
// C++ implementation to find the sum of all the // terms in the nth row of the given series #include <bits/stdc++.h> using namespace std;
// function to find the required sum int sumOfTermsInNthRow( int n)
{ // sum = n * (2 * n^2 + 1)
int sum = n * (2 * pow (n, 2) + 1);
return sum;
} // Driver program to test above int main()
{ int n = 4;
cout << "Sum of all the terms in nth row = "
<< sumOfTermsInNthRow(n);
return 0;
} |
// Java implementation to find the sum of all the // terms in the nth row of the given series import static java.lang.Math.pow;
class Test {
// method to find the required sum
static int sumOfTermsInNthRow( int n)
{
// sum = n * (2 * n^2 + 1)
int sum = ( int )(n * ( 2 * pow(n, 2 ) + 1 ));
return sum;
}
// Driver method
public static void main(String args[])
{
int n = 4 ;
System.out.println( "Sum of all the terms in nth row = "
+ sumOfTermsInNthRow(n));
}
} |
# Python 3 implementation to find # the sum of all the terms in the # nth row of the given series from math import pow
# function to find the required sum def sumOfTermsInNthRow(n):
# sum = n * (2 * n^2 + 1)
sum = n * ( 2 * pow (n, 2 ) + 1 )
return sum
# Driver Code if __name__ = = '__main__' :
n = 4
print ( "Sum of all the terms in nth row =" ,
int (sumOfTermsInNthRow(n)))
# This code is contributed # by Surendra_Gangwar |
// C# implementation to find the sum of all the // terms in the nth row of the given series using System;
class Test {
// method to find the required sum
static int sumOfTermsInNthRow( int n)
{
// sum = n * (2 * n^2 + 1)
int sum = ( int )(n * (2 * Math.Pow(n, 2) + 1));
return sum;
}
// Driver method
public static void Main()
{
int n = 4;
Console.Write( "Sum of all the terms in nth row = "
+ sumOfTermsInNthRow(n));
}
} // This code is contributed by vt_m. |
<?php // PHP implementation to find // the sum of all the terms in // the nth row of the given series // function to find the required sum function sumOfTermsInNthRow( $n )
{ // sum = n * (2 * n^2 + 1)
$sum = $n * (2 * pow( $n , 2) + 1);
return $sum ;
} // Driver Code
$n = 4;
echo "Sum of all the terms in nth row = " ,
sumOfTermsInNthRow( $n );
// This code is contributed by ajit ?> |
<script> // javascript implementation to find the sum of all the // terms in the nth row of the given series // function to find the required sum function sumOfTermsInNthRow( n)
{ // sum = n * (2 * n^2 + 1)
let sum = n * (2 * Math.pow(n, 2) + 1);
return sum;
} // Driver program to test above let n = 4;
document.write( "Sum of all the terms in nth row = "
+ sumOfTermsInNthRow(n));
// This code is contributed by aashish1995 </script> |
Output:
Sum of all the terms in nth row = 132
Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.