Sum of the series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n)
Last Updated :
11 Aug, 2022
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++
#include <bits/stdc++.h>
using namespace std;
int sumOfTheSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++) {
int k = 2;
for ( int j = 1; j <= i; j++) {
sum += k;
k += 2;
}
}
return sum;
}
int main()
{
int n = 5;
cout << "Sum = "
<< sumOfTheSeries(n);
return 0;
}
|
Java
class GFG{
static int sumOfTheSeries( int n)
{
int sum = 0 ;
for ( int i = 1 ; i <= n; i++) {
int k = 2 ;
for ( int j = 1 ; j <= i; j++) {
sum += k;
k += 2 ;
}
}
return sum;
}
public static void main(String[] args)
{
int n = 5 ;
System.out.printf( "Sum = %d" ,
sumOfTheSeries(n));
}
}
|
Python3
def sumOfTheSeries(n):
sum = 0
for i in range ( 0 , n + 1 ):
k = 2
for j in range ( 1 , i + 1 ):
sum = sum + k;
k = k + 2
return sum ;
n = 5
ans = sumOfTheSeries(n);
print (ans)
|
C#
using System;
class GFG{
static int sumOfTheSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++) {
int k = 2;
for ( int j = 1; j <= i; j++) {
sum += k;
k += 2;
}
}
return sum;
}
public static void Main()
{
int n = 5;
Console.Write( "Sum = " +
sumOfTheSeries(n));
}
}
|
PHP
<?php
function sumOfTheSeries( $n )
{
$sum = 0;
for ( $i = 1; $i <= $n ; $i ++)
{
$k = 2;
for ( $j = 1; $j <= $i ; $j ++)
{
$sum += $k ;
$k += 2;
}
}
return $sum ;
}
$n = 5;
echo "Sum = " , sumOfTheSeries( $n );
?>
|
Javascript
<script>
function sumOfTheSeries(n)
{
let sum = 0;
for (let i = 1; i <= n; i++)
{
let k = 2;
for (let j = 1; j <= i; j++)
{
sum += k;
k += 2;
}
}
return sum;
}
let n = 5;
document.write( "Sum = " + sumOfTheSeries(n));
</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++
#include <bits/stdc++.h>
using namespace std;
int sumOfTheSeries( int n)
{
int sum_n = (n * (n + 1) / 2);
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
return (sum_n + sum_sq_n);
}
int main()
{
int n = 5;
cout << "Sum = "
<< sumOfTheSeries(n);
return 0;
}
|
Java
class GFG{
static int sumOfTheSeries( int n)
{
int sum_n = (n * (n + 1 ) / 2 );
int sum_sq_n = (n * (n + 1 ) / 2 ) *
( 2 * n + 1 ) / 3 ;
return (sum_n + sum_sq_n);
}
public static void main(String[] args)
{
int n = 5 ;
System.out.printf( "Sum = %d" ,
sumOfTheSeries(n));
}
}
|
Python3
def sumOfTheSeries(n):
sum_n = int ((n * (n + 1 ) / 2 ));
sum_sq_n = int ((n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 )
return (sum_n + sum_sq_n);
n = 5
ans = sumOfTheSeries(n)
print (ans)
|
C#
using System;
class GFG{
static int sumOfTheSeries( int n)
{
int sum_n = (n * (n + 1) / 2);
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
return (sum_n + sum_sq_n);
}
public static void Main()
{
int n = 5;
Console.Write( "Sum = " +
sumOfTheSeries(n));
}
}
|
PHP
<?php
function sumOfTheSeries( $n )
{
$sum_n = ( $n * ( $n + 1) / 2);
$sum_sq_n = ( $n * ( $n + 1) / 2) *
(2 * $n + 1) / 3;
return ( $sum_n +
$sum_sq_n );
}
$n = 5;
echo ( "Sum = " .sumOfTheSeries( $n ));
?>
|
Javascript
<script>
function sumOfTheSeries(n)
{
let sum_n = (n * (n + 1) / 2);
let sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
return (sum_n + sum_sq_n);
}
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)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...