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)
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