Sum of the series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n)
Given a positive integer n. The problem is to find the sum of the given series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n), where i-th term in the series is the sum of first i even natural numbers.
Examples:
Input : n = 2 Output : 8 (2) + (2+4) = 8 Input : n = 5 Output : 70 (2) + (2+4) + (2+4+6) + (2+4+6+8) + (2+4+6+8+10) = 70
Naive Approach: Using two loops get the sum of each i-th term and then add those sum to the final sum.
C++
// C++ implementation to find the sum // of the given series #include <bits/stdc++.h> using namespace std; // function to find the sum // of the given series int sumOfTheSeries( int n) { int sum = 0; for ( int i = 1; i <= n; i++) { // first term of each i-th term int k = 2; for ( int j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } // Driver program to test above int main() { int n = 5; cout << "Sum = " << sumOfTheSeries(n); return 0; } |
Java
// Java implementation to find the // sum of the given series class GFG{ // function to find the sum // of the given series static int sumOfTheSeries( int n) { int sum = 0 ; for ( int i = 1 ; i <= n; i++) { // first term of each i-th term int k = 2 ; for ( int j = 1 ; j <= i; j++) { sum += k; // next term k += 2 ; } } // required sum return sum; } // Driver program to test above public static void main(String[] args) { int n = 5 ; System.out.printf( "Sum = %d" , sumOfTheSeries(n)); } } // This code is contributed by // Smitha Dinesh Semwal |
Python3
# Python3 implementation to find # the sum of the given series # function to find the sum # of the given series def sumOfTheSeries(n): sum = 0 for i in range ( 0 , n + 1 ): # first term of each i-th # term k = 2 for j in range ( 1 , i + 1 ): sum = sum + k; # next term k = k + 2 # required sum return sum ; # Driver program to test above n = 5 ans = sumOfTheSeries(n); print (ans) # This code is contributed by saloni1297. |
C#
// C# implementation to find the // sum of the given series using System; class GFG{ // function to find the sum // of the given series static int sumOfTheSeries( int n) { int sum = 0; for ( int i = 1; i <= n; i++) { // first term of each i-th term int k = 2; for ( int j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } // Driver program to test above public static void Main() { int n = 5; Console.Write( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by // vt_m |
PHP
<?php // PHP implementation to find // the sum of the given series // function to find the sum // of the given series function sumOfTheSeries( $n ) { $sum = 0; for ( $i = 1; $i <= $n ; $i ++) { // first term of each // i-th term $k = 2; for ( $j = 1; $j <= $i ; $j ++) { $sum += $k ; // next term $k += 2; } } // required sum return $sum ; } // Driver program to test above $n = 5; echo "Sum = " , sumOfTheSeries( $n ); // This code is contributed by ajit. ?> |
Javascript
<script> // Javascript implementation to find // the sum of the given series // function to find the sum // of the given series function sumOfTheSeries(n) { let sum = 0; for (let i = 1; i <= n; i++) { // first term of each // i-th term let k = 2; for (let j = 1; j <= i; j++) { sum += k; // next term k += 2; } } // required sum return sum; } // Driver program to test above let n = 5; document.write( "Sum = " + sumOfTheSeries(n)); // This code is contributed by gfgking. </script> |
Output:
Sum = 70
Efficient Approach:
Let an be the n-th term of the given series.
an = (2 + 4 + 6 + 8 +....+ 2n). = sum of first n even numbers. = n * (n + 1). = n2 + n.
Refer this post for the proof of above formula.
Now,
Refer this and this post for the proof of above formula.
C++
// C++ implementation to find the sum // of the given series #include <bits/stdc++.h> using namespace std; // function to find the sum // of the given series int sumOfTheSeries( int n) { // sum of 1st n natural numbers int sum_n = (n * (n + 1) / 2); // sum of squares of 1st n natural numbers int sum_sq_n = (n * (n + 1) / 2) * (2 * n + 1) / 3; // required sum return (sum_n + sum_sq_n); } // Driver program to test above int main() { int n = 5; cout << "Sum = " << sumOfTheSeries(n); return 0; } |
Java
// Java implementation to find the // sum of the given series class GFG{ // function to find the sum // of the given series static int sumOfTheSeries( int n) { // sum of 1st n natural numbers int sum_n = (n * (n + 1 ) / 2 ); // sum of squares of 1st n natural // numbers int sum_sq_n = (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ; // required sum return (sum_n + sum_sq_n); } // Driver program to test above public static void main(String[] args) { int n = 5 ; System.out.printf( "Sum = %d" , sumOfTheSeries(n)); } } // This code is contributed by //Smitha Dinesh Semwal |
Python3
# Python3 implementation to find # the sum of the given series # function to find the sum # of the given series def sumOfTheSeries(n): # sum of 1st n natural numbers sum_n = int ((n * (n + 1 ) / 2 )); # sum of squares of 1st n natural numbers sum_sq_n = int ((n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ) # required sum return (sum_n + sum_sq_n); # Driver program to test above n = 5 ans = sumOfTheSeries(n) print (ans) # This code is contributed by saloni1297. |
C#
// C# implementation to find the // sum of the given series using System; class GFG{ // function to find the sum // of the given series static int sumOfTheSeries( int n) { // sum of 1st n natural numbers int sum_n = (n * (n + 1) / 2); // sum of squares of 1st n // natural numbers int sum_sq_n = (n * (n + 1) / 2) * (2 * n + 1) / 3; // required sum return (sum_n + sum_sq_n); } // Driver program to test above public static void Main() { int n = 5; Console.Write( "Sum = " + sumOfTheSeries(n)); } } // This code is contributed by // vt_m |
PHP
<?php // PHP implementation // to find the sum // of the given series // function to find the // sum of the given series function sumOfTheSeries( $n ) { // sum of 1st n // natural numbers $sum_n = ( $n * ( $n + 1) / 2); // sum of squares of // 1st n natural numbers $sum_sq_n = ( $n * ( $n + 1) / 2) * (2 * $n + 1) / 3; // required sum return ( $sum_n + $sum_sq_n ); } // Driver Code $n = 5; echo ( "Sum = " .sumOfTheSeries( $n )); // This code is contributed by // Manish Shaw(manishshaw1) ?> |
Javascript
<script> // JavaScript Program to find the // sum of the given series // function to find the sum // of the given series function sumOfTheSeries(n) { // sum of 1st n natural numbers let sum_n = (n * (n + 1) / 2); // sum of squares of 1st n natural // numbers let sum_sq_n = (n * (n + 1) / 2) * (2 * n + 1) / 3; // required sum return (sum_n + sum_sq_n); } // Driver code let n = 5; document.write( "Sum = " + sumOfTheSeries(n)); </script> |
Output:
Sum = 70
Time complexity: O(1) as it is performing constant operations
Auxiliary space: O(1)