Question

Write python program for the sum of first n even numbers

Answer 1

// C++ implementation to find sum of
// first n even numbers
#include <bits/stdc++.h>



using namespace std;


// function to find sum of
// first n even numbers

int evenSum(int n)
{

int curr = 2, sum = 0;



// sum of first n even numbers

for (int i = 1; i <= n; i++) {

sum += curr;



// next even number

curr += 2;

}



// required sum

return sum;
}


// Driver program to test above

int main()
{

int n = 20;
Proof:

Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.

Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)
// C++ implementation to find sum of
// first n even numbers
#include <bits/stdc++.h>

using namespace std;


// function to find sum of
// first n even numbers

int evenSum(int n)
{

// required sum

return (n * (n + 1));
}


// Driver program to test above

int main()
{

int n = 20;

cout << "Sum of first " << n

<< " Even numbers is: " << evenSum(n);

return 0;
}

Output:
Sum of first 20 Even numbers is: 420

cout << "Sum of first " << n

<< " Even numbers is: " << evenSum(n);

return 0;
}

Answer 2

1. Make a variable to hold the running total

2. Iterate through each number less than or equal to n

    - add each number you iterate through to the variable

3. return the sum variable

   def sumFirstN(n):

       sum = 0

       for i in range(n+1):

           sum = sum + i

       return sum

   print(firstN(10))n)

Note: you need "for i in range(n+1):" because range generates a list of numbers up to (but not including) n, so we need to pass it n+1 to include n