Program to find sum of series 1 + 2 + 2 + 3 + 3 + 3 + . . . + n
Given a positive integer n and the task is to find sum of series 1 + 2 + 2 + 3 + 3 + 3 + . . . + n.
Examples:
Input : n = 5
Output : 55
= 1 + 2 + 2 + 3 + 3 + 3 + 4 + 4 + 4 +
4 + 5 + 5 + 5 + 5 + 5.
= 55
Input : n = 10
Output : 385
Addition method: In addition method sum all the elements one by one.
Below is the implementation of this approach.
C++
#include <bits/stdc++.h>
using namespace std;
int sumOfSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++)
for ( int j = 1; j <= i; j++)
sum = sum + i;
return sum;
}
int main()
{
int n = 10;
cout << sumOfSeries(n);
return 0;
}
|
Java
public class GfG{
static int sumOfSeries( int n)
{
int sum = 0 ;
for ( int i = 1 ; i <= n; i++)
for ( int j = 1 ; j <= i; j++)
sum = sum + i;
return sum;
}
public static void main(String s[])
{
int n = 10 ;
System.out.println(sumOfSeries(n));
}
}
|
Python3
import math
def sumOfSeries( n):
sum = 0
for i in range ( 1 , n + 1 ):
sum = sum + i * i
return sum
n = 10
print (sumOfSeries(n))
|
C#
using System;
public class GfG {
static int sumOfSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++)
for ( int j = 1; j <= i; j++)
sum = sum + i;
return sum;
}
public static void Main()
{
int n = 10;
Console.Write(sumOfSeries(n));
}
}
|
PHP
<?php
function sumOfSeries( $n )
{
$sum = 0;
for ( $i = 1; $i <= $n ; $i ++)
for ( $j = 1; $j <= $i ; $j ++)
$sum = $sum + $i ;
return $sum ;
}
$n = 10;
echo (sumOfSeries( $n ));
?>
|
Javascript
<script>
function sumOfSeries( n) {
let sum = 0;
for (let i = 1; i <= n; i++)
for (let j = 1; j <= i; j++)
sum = sum + i;
return sum;
}
let n = 10;
document.write(sumOfSeries(n));
</script>
|
Output:
385
Time Complexity: O(n2)
Auxiliary Space: O(1)
Multiplication method:In multiplication method every elements multiply by itself and then add them.
Input n = 10
sum = 1 + 2 + 2 + 3 + 3 + 3 + 4 + . . . + 10
= 1 + 2 * 2 + 3 * 3 + 4 * 4 + . . . + 10 * 10
= 1 + 4 + 9 + 16 + . . . + 100
= 385
C++
#include <bits/stdc++.h>
using namespace std;
int sumOfSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++)
sum = sum + i * i;
return sum;
}
int main()
{
int n = 10;
cout << sumOfSeries(n);
return 0;
}
|
Java
public class GfG{
static int sumOfSeries( int n)
{
int sum = 0 ;
for ( int i = 1 ; i <= n; i++)
sum = sum + i * i;
return sum;
}
public static void main(String args[])
{
int n = 10 ;
System.out.println(sumOfSeries(n));
}
}
|
Python3
import math
def sumOfSeries( n):
sum = 0
for i in range ( 1 , n + 1 ):
sum = sum + i * i
return sum
n = 10
print (sumOfSeries(n))
|
C#
using System;
class GfG {
static int sumOfSeries( int n)
{
int sum = 0;
for ( int i = 1; i <= n; i++)
sum = sum + i * i;
return sum;
}
public static void Main()
{
int n = 10;
Console.WriteLine(sumOfSeries(n));
}
}
|
PHP
<?php
function sumOfSeries( $n )
{
$sum = 0;
for ( $i = 1; $i <= $n ; $i ++)
$sum = $sum + $i * $i ;
return $sum ;
}
$n = 10;
echo (sumOfSeries( $n ));
?>
|
Javascript
<script>
function sumOfSeries(n)
{
var sum = 0;
for (let i = 1; i <= n; i++)
sum = sum + i * i;
return sum;
}
var n = 10;
document.write(sumOfSeries(n));
</script>
|
Output:
385
Time Complexity: O(n)
Auxiliary Space: O(1)
Using formula: We also use formula to find the sum of series.
Input n = 10;
Sum of series = (n * (n + 1) * (2 * n + 1)) / 6
put n = 10 in the above formula
sum = (10 * (10 + 1) * (2 * 10 + 1)) / 6
= (10 * 11 * 21) / 6
= 385
C++
#include <bits/stdc++.h>
using namespace std;
int sumOfSeries( int n)
{
return (n * (n + 1) * (2 * n + 1)) / 6;
}
int main()
{
int n = 10;
cout << sumOfSeries(n);
return 0;
}
|
Java
public class GfG
{
static int sumOfSeries( int n)
{
return (n * (n + 1 ) * ( 2 * n + 1 )) / 6 ;
}
public static void main(String s[])
{
int n = 10 ;
System.out.println(sumOfSeries(n));
}
}
|
Python3
import math
def sumOfSeries( n):
return ((n * (n + 1 ) * ( 2 * n + 1 )) / 6 )
n = 10
print (sumOfSeries(n))
|
C#
using System;
public class GfG {
static int sumOfSeries( int n)
{
return (n * (n + 1) * (2 * n + 1)) / 6;
}
public static void Main()
{
int n = 10;
Console.WriteLine(sumOfSeries(n));
}
}
|
PHP
<?php
function sumOfSeries( $n )
{
return ( $n * ( $n + 1) *
(2 * $n + 1)) / 6;
}
$n = 10;
echo (sumOfSeries( $n ));
?>
|
Javascript
<script>
function sumOfSeries(n)
{
return (n * (n + 1) * (2 * n + 1)) / 6;
}
var n = 10;
document.write(sumOfSeries(n));
</script>
|
Output :
385
Time Complexity: O(1)
Auxiliary Space: O(1)
Please refer sum of squares of natural numbers for details of above formula and more optimizations.
Last Updated :
30 Nov, 2021
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