You have been given a series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n), find out the sum of the series till nth term.
Examples :
Input : n = 3
Output : 14
Explanation : 1 + 1/2^2 + 1/3^3
Input : n = 5
Output : 55
Explanation : (1*1) + (2*2) + (3*3) + (4*4) + (5*5)
C++
#include<iostream>
using namespace std;
int Series(int n)
{
int i;
int sums = 0;
for (i = 1; i <= n; i++)
sums += (i * i);
return sums;
}
int main()
{
int n = 3;
int res = Series(n);
cout<<res<<endl;
}
|
C
#include <stdio.h>
int Series(int n)
{
int i;
int sums = 0;
for (i = 1; i <= n; i++)
sums += (i * i);
return sums;
}
int main()
{
int n = 3;
int res = Series(n);
printf("%d", res);
}
|
Java
import java.io.*;
class GFG {
static int Series(int n)
{
int i;
int sums = 0;
for (i = 1; i <= n; i++)
sums += (i * i);
return sums;
}
public static void main(String[] args)
{
int n = 3;
int res = Series(n);
System.out.println(res);
}
}
|
Python
def Series(n):
sums = 0
for i in range(1, n + 1):
sums += (i * i);
return sums
n = 3
res = Series(n)
print(res)
|
C#
using System;
class GFG {
static int Series(int n)
{
int i;
int sums = 0;
for (i = 1; i <= n; i++)
sums += (i * i);
return sums;
}
public static void Main()
{
int n = 3;
int res = Series(n);
Console.Write(res);
}
}
|
PHP
<?php
function Series($n)
{
$i;
$sums = 0;
for ($i = 1; $i <= $n; $i++)
$sums += ($i * $i);
return $sums;
}
$n = 3;
$res = Series($n);
echo($res);
?>
|
Javascript
<script>
function Series( n) {
let i;
let sums = 0;
for (i = 1; i <= n; i++)
sums += (i * i);
return sums;
}
let n = 3;
let res = Series(n);
document.write(res);
</script>
|
Output :
14
Time Complexity: O(n)
Please refer below post for O(1) solution.
Sum of squares of first n natural numbers
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