Given the value of the n. You have to find the sum of the series where the nth term of the sequence is given by:
Tn = n2 – ( n – 1 )2
Examples :
Input : 3
Output : 9
Explanation: So here the term of the sequence upto n = 3 are:
1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9
Input : 6
Output : 36Simple Approach
Just use a loop and calculate the sum of each term and print the sum.
C++
#include <bits/stdc++.h>
using namespace std;
int summingSeries(long n)
{
int S = 0;
for (int i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
int main()
{
int n = 100;
cout << "The sum of n term is: "
<< summingSeries(n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
class GFG
{
static int summingSeries(long n)
{
int S = 0;
for (i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
public static void main(String[] args)
{
int n = 100;
System.out.println("The sum of n term is: " +
summingSeries(n));
}
}
|
Python3
def summingSeries(n):
S = 0
for i in range(1, n+1):
S += i * i - (i - 1) * (i - 1)
return S
n = 100
print("The sum of n term is: ",
summingSeries(n), sep = "")
|
C#
using System;
class GFG
{
static int summingSeries(long n)
{
return (int)Math.Pow(n, 2);
}
public static void Main(String[] args)
{
int n = 100;
Console.Write("The sum of n term is: " +
summingSeries(n));
}
}
|
PHP
<?php
function summingSeries( $n)
{
$S = 0;
for ($i = 1; $i <= $n; $i++)
$S += $i * $i - ($i - 1) *
($i - 1);
return $S;
}
$n = 100;
echo "The sum of n term is: ",
summingSeries($n) ;
?>
|
Javascript
<script>
function summingSeries(n)
{
let S = 0;
for (let i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
let n = 100;
document.write("The sum of n term is: " + summingSeries(n));
</script>
|
Output:
The sum of n term is: 10000
Time complexity – O(N)
Space complexity – O(1)
Efficient Approach
Use of mathematical approach can solve this problem in more efficient way.
Tn = n2 – (n-1)2
Sum of the series is given by (S) = SUM( Tn )
LET US TAKE A EXAMPLE IF
N = 4
It means there should be 4 terms in the series so
1st term = 12 – ( 1 – 1 )2
2nd term = 22 – ( 2 – 1 )2
3th term = 32 – ( 3 – 1 )2
4th term = 42 – ( 3 – 1 )2
SO SUM IS GIVEN BY = (1 – 0) + (4 – 1) + (9 – 4) + (16 – 9)
= 16
FROM THIS WE HAVE NOTICE THAT 1, 4, 9 GET CANCELLED FROM THE SERIES
ONLY 16 IS LEFT WHICH IS EQUAL TO THE SQUARE OF N
So from the above series we notice that each term gets canceled from the next term, only the last term is left which is equal to N2.
C++
#include <bits/stdc++.h>
using namespace std;
int summingSeries(long n)
{
return pow(n, 2);
}
int main()
{
int n = 100;
cout << "The sum of n term is: "
<< summingSeries(n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
class GFG
{
static int summingSeries(long n)
{
return (int)Math.pow(n, 2);
}
public static void main(String[] args)
{
int n = 100;
System.out.println("The sum of n term is: " +
summingSeries(n));
}
}
|
Python3
import math
def summingSeries(n):
return math.pow(n, 2)
n = 100
print ("The sum of n term is: ",
summingSeries(n))
|
C#
using System;
class GFG
{
static int summingSeries(long n)
{
return (int)Math.Pow(n, 2);
}
public static void Main()
{
int n = 100;
Console.Write("The sum of n term is: " +
summingSeries(n));
}
}
|
PHP
<?php
function summingSeries($n)
{
return pow($n, 2);
}
$n = 100;
echo "The sum of n term is: ",
summingSeries($n);
?>
|
Javascript
<script>
function summingSeries(n)
{
return Math.pow(n, 2);
}
let n = 100;
document.write("The sum of n term is: "
+ summingSeries(n) + "<br>");
</script>
|
Output:
The sum of n term is: 10000
Time complexity – O(1)
Space complexity – O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
