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)
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