Given a mathematical series as 3, 9, 21, 41, 71… For a given integer n, you have to find the nth number of this series.
Examples :
Input : n = 4 Output : 41 Input : n = 2 Output : 9
Our first task for solving this problem is to crack the series. If you will gave a closer look on the series, for a general n-th term the value will be (?n2)+(?n)+1, where
- (?n2): is sum of squares of first n-natural numbers.
- (?n): is sum of first n-natural numbers.
- 1 is a simple constant value.
So, for calculating any n-th term of given series say f(n) we have:
f(n) = (?n2)+(?n)+1
= ( ((n*(n+1)*(2n+1))/6) + (n*(n+1)/2) + 1
= (n3+ 3n2 + 2n + 3 ) /3
C++
// Program to calculate // nth term of a series #include <bits/stdc++.h> using namespace std;
// func for calualtion int seriesFunc( int n)
{ // for summation of square
// of first n-natural nos.
int sumSquare = (n * (n + 1)
* (2 * n + 1)) / 6;
// summation of first n natural nos.
int sumNatural = (n * (n + 1) / 2);
// return result
return (sumSquare + sumNatural + 1);
} // Driver Code int main()
{ int n = 8;
cout << seriesFunc(n) << endl;
n = 13;
cout << seriesFunc(13);
return 0;
} |
Java
// Java Program to calculate // nth term of a series import java.io.*;
class GFG
{ // func for calualtion
static int seriesFunc( int n)
{
// for summation of square
// of first n-natural nos.
int sumSquare = (n * (n + 1 )
* ( 2 * n + 1 )) / 6 ;
// summation of first n natural nos.
int sumNatural = (n * (n + 1 ) / 2 );
// return result
return (sumSquare + sumNatural + 1 );
}
// Driver Code
public static void main(String args[])
{
int n = 8 ;
System.out.println(seriesFunc(n));
n = 13 ;
System.out.println(seriesFunc( 13 ));
}
} // This code is contributed by Nikita Tiwari. |
Python3
# Program to calculate # nth term of a series # func for calualtion def seriesFunc(n):
# for summation of square
# of first n-natural nos.
sumSquare = (n * (n + 1 ) *
( 2 * n + 1 )) / 6
# summation of first n
# natural nos.
sumNatural = (n * (n + 1 ) / 2 )
# return result
return (sumSquare + sumNatural + 1 )
# Driver Code n = 8
print ( int (seriesFunc(n)))
n = 13
print ( int (seriesFunc(n)))
# This is code is contributed by Shreyanshi Arun. |
C#
// C# program to calculate // nth term of a series using System;
class GFG
{ // Function for calualtion
static int seriesFunc( int n)
{
// For summation of square
// of first n-natural nos.
int sumSquare = (n * (n + 1)
* (2 * n + 1)) / 6;
// summation of first n natural nos.
int sumNatural = (n * (n + 1) / 2);
// return result
return (sumSquare + sumNatural + 1);
}
// Driver Code
public static void Main()
{
int n = 8;
Console.WriteLine(seriesFunc(n));
n = 13;
Console.WriteLine(seriesFunc(13));
}
} // This code is contributed by vt_m. |
PHP
<?php // Program to calculate // nth term of a series // func for calualtion function seriesFunc( $n )
{ // for summation of square
// of first n-natural nos.
$sumSquare = ( $n * ( $n + 1)
* (2 * $n + 1)) / 6;
// summation of first n natural nos.
$sumNatural = ( $n * ( $n + 1) / 2);
// return result
return ( $sumSquare + $sumNatural + 1);
} // Driver Code $n = 8;
echo (seriesFunc( $n ) . "\n" );
$n = 13;
echo (seriesFunc( $n ) . "\n" );
// This code is contributed by Ajit. ?> |
Javascript
<script> // JavaScript Program to calculate // nth term of a series // func for calualtion
function seriesFunc(n)
{
// for summation of square
// of first n-natural nos.
let sumSquare = (n * (n + 1)
* (2 * n + 1)) / 6;
// summation of first n natural nos.
let sumNatural = (n * (n + 1) / 2);
// return result
return (sumSquare + sumNatural + 1);
}
// Driver code let n = 8;
document.write(seriesFunc(n) + "<br/>" );
n = 13;
document.write(seriesFunc(13));
</script> |
Output :
241 911
Time Complexity: O(1) since constant operations are performed
Auxiliary Space: O(1)