Given three integers, A, X and n, the task is to print terms of below binomial expression series.
(A+X)n = nC0AnX0 + nC1An-1X1 + nC2An-2X2 +….+ nCnA0Xn
Examples:
Input : A = 1, X = 1, n = 5 Output : 1 5 10 10 5 1 Input : A = 1, B = 2, n = 6 Output : 1 12 60 160 240 192 64
Simple Solution : We know that for each value of n there will be (n+1) term in the binomial series. So now we use a simple approach and calculate the value of each element of the series and print it .
nCr = (n!) / ((n-r)! * (r)!) Below is value of general term. Tr+1 = nCn-rAn-rXr So at each position we have to find the value of the general term and print that term .
// CPP program to print terms of binomial // series and also calculate sum of series. #include <bits/stdc++.h> using namespace std;
// function to calculate factorial of // a number int factorial( int n)
{ int f = 1;
for ( int i = 2; i <= n; i++)
f *= i;
return f;
} // function to print the series void series( int A, int X, int n)
{ // calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for ( int i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = pow (A, n - i);
int xPow = pow (X, i);
// display the series
cout << (nFact * aPow * xPow) /
(niFact * iFact) << " " ;
}
} // main function started int main()
{ int A = 3, X = 4, n = 5;
series(A, X, n);
return 0;
} |
// Java program to print terms of binomial // series and also calculate sum of series. import java.io.*;
class GFG {
// function to calculate factorial of
// a number
static int factorial( int n)
{
int f = 1 ;
for ( int i = 2 ; i <= n; i++)
f *= i;
return f;
}
// function to print the series
static void series( int A, int X, int n)
{
// calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for ( int i = 0 ; i < n + 1 ; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = ( int )Math.pow(A, n - i);
int xPow = ( int )Math.pow(X, i);
// display the series
System.out.print((nFact * aPow * xPow)
/ (niFact * iFact) + " " );
}
}
// main function started
public static void main(String[] args)
{
int A = 3 , X = 4 , n = 5 ;
series(A, X, n);
}
} // This code is contributed by vt_m. |
# Python3 program to print terms of binomial # series and also calculate sum of series. # function to calculate factorial # of a number def factorial(n):
f = 1
for i in range ( 2 , n + 1 ):
f * = i
return f
# Function to print the series def series(A, X, n):
# calculating the value of n!
nFact = factorial(n)
# loop to display the series
for i in range ( 0 , n + 1 ):
# For calculating the
# value of nCr
niFact = factorial(n - i)
iFact = factorial(i)
# calculating the value of
# A to the power k and X to
# the power k
aPow = pow (A, n - i)
xPow = pow (X, i)
# display the series
print ( int ((nFact * aPow * xPow) /
(niFact * iFact)), end = " " )
# Driver Code A = 3 ; X = 4 ; n = 5
series(A, X, n) # This code is contributed by Smitha Dinesh Semwal. |
// C# program to print terms of binomial // series and also calculate sum of series. using System;
class GFG {
// function to calculate factorial of
// a number
static int factorial( int n)
{
int f = 1;
for ( int i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
static void series( int A, int X, int n)
{
// calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for ( int i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = ( int )Math.Pow(A, n - i);
int xPow = ( int )Math.Pow(X, i);
// display the series
Console.Write((nFact * aPow * xPow)
/ (niFact * iFact) + " " );
}
}
// main function started
public static void Main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
} // This code is contributed by anuj_67. |
<?php // PHP program to print // terms of binomial // series and also // calculate sum of series. // function to calculate // factorial of a number function factorial( $n )
{ $f = 1;
for ( $i = 2; $i <= $n ; $i ++)
$f *= $i ;
return $f ;
} // function to print the series function series( $A , $X , $n )
{ // calculating the
// value of n!
$nFact = factorial( $n );
// loop to display
// the series
for ( $i = 0; $i < $n + 1; $i ++)
{
// For calculating the
// value of nCr
$niFact = factorial( $n - $i );
$iFact = factorial( $i );
// calculating the value of
// A to the power k and X to
// the power k
$aPow = pow( $A , $n - $i );
$xPow = pow( $X , $i );
// display the series
echo ( $nFact * $aPow * $xPow ) /
( $niFact * $iFact ) , " " ;
}
} // Driver Code
$A = 3;
$X = 4;
$n = 5;
series( $A , $X , $n );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript program to print terms of binomial // series and also calculate sum of series. // function to calculate factorial of
// a number
function factorial(n)
{
let f = 1;
for (let i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
function series(A, X, n)
{
// calculating the value of n!
let nFact = factorial(n);
// loop to display the series
for (let i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
let niFact = factorial(n - i);
let iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
let aPow = Math.pow(A, n - i);
let xPow = Math.pow(X, i);
// display the series
document.write((nFact * aPow * xPow)
/ (niFact * iFact) + " " );
}
}
// Driver Code let A = 3, X = 4, n = 5;
series(A, X, n);
// This code is contributed by chinmoy1997pal.
</script> |
243 1620 4320 5760 3840 1024
Time complexity : O(n2)
Auxiliary Space : O(1)
Efficient Solution :
The idea is to compute next term using previous term. We can compute next term in O(1) time. We use below property of Binomial Coefficients.
nCi+1 = nCi*(n-i)/(i+1)
// CPP program to print terms of binomial // series and also calculate sum of series. #include <bits/stdc++.h> using namespace std;
// function to print the series void series( int A, int X, int n)
{ // Calculating and printing first term
int term = pow (A, n);
cout << term << " " ;
// Computing and printing remaining terms
for ( int i = 1; i <= n; i++) {
// Find current term using previous terms
// We increment power of X by 1, decrement
// power of A by 1 and compute nCi using
// previous term by multiplying previous
// term with (n - i + 1)/i
term = term * X * (n - i + 1)/(i * A);
cout << term << " " ;
}
} // main function started int main()
{ int A = 3, X = 4, n = 5;
series(A, X, n);
return 0;
} |
// Java program to print terms of binomial // series and also calculate sum of series. import java.io.*;
class GFG {
// function to print the series
static void series( int A, int X, int n)
{
// Calculating and printing first
// term
int term = ( int )Math.pow(A, n);
System.out.print(term + " " );
// Computing and printing
// remaining terms
for ( int i = 1 ; i <= n; i++) {
// Find current term using
// previous terms We increment
// power of X by 1, decrement
// power of A by 1 and compute
// nCi using previous term by
// multiplying previous term
// with (n - i + 1)/i
term = term * X * (n - i + 1 )
/ (i * A);
System.out.print(term + " " );
}
}
// main function started
public static void main(String[] args)
{
int A = 3 , X = 4 , n = 5 ;
series(A, X, n);
}
} // This code is contributed by vt_m. |
# Python 3 program to print terms of binomial # series and also calculate sum of series. # Function to print the series def series(A, X, n):
# Calculating and printing first term
term = pow (A, n)
print (term, end = " " )
# Computing and printing remaining terms
for i in range ( 1 , n + 1 ):
# Find current term using previous terms
# We increment power of X by 1, decrement
# power of A by 1 and compute nCi using
# previous term by multiplying previous
# term with (n - i + 1)/i
term = int (term * X * (n - i + 1 ) / (i * A))
print (term, end = " " )
# Driver Code A = 3 ; X = 4 ; n = 5
series(A, X, n) # This code is contributed by Smitha Dinesh Semwal. |
// C# program to print terms of binomial // series and also calculate sum of series. using System;
public class GFG {
// function to print the series
static void series( int A, int X, int n)
{
// Calculating and printing first
// term
int term = ( int )Math.Pow(A, n);
Console.Write(term + " " );
// Computing and printing
// remaining terms
for ( int i = 1; i <= n; i++) {
// Find current term using
// previous terms We increment
// power of X by 1, decrement
// power of A by 1 and compute
// nCi using previous term by
// multiplying previous term
// with (n - i + 1)/i
term = term * X * (n - i + 1)
/ (i * A);
Console.Write(term + " " );
}
}
// main function started
public static void Main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
} // This code is contributed by anuj_67. |
<?php // PHP program to print // terms of binomial // series and also // calculate sum of // series. // function to print // the series function series( $A , $X , $n )
{ // Calculating and printing
// first term
$term = pow( $A , $n );
echo $term , " " ;
// Computing and printing
// remaining terms
for ( $i = 1; $i <= $n ; $i ++)
{
// Find current term
// using previous terms
// We increment power
// of X by 1, decrement
// power of A by 1 and
// compute nCi using
// previous term by
// multiplying previous
// term with (n - i + 1)/i
$term = $term * $X * ( $n - $i + 1) /
( $i * $A );
echo $term , " " ;
}
} // Driver Code
$A = 3;
$X = 4;
$n = 5;
series( $A , $X , $n );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript program to print terms of binomial // series and also calculate sum of series. // function to print the series function series(A, X, n)
{ // Calculating and printing first term
let term = Math.pow(A, n);
document.write(term + " " );
// Computing and printing remaining terms
for (let i = 1; i <= n; i++) {
// Find current term using previous terms
// We increment power of X by 1, decrement
// power of A by 1 and compute nCi using
// previous term by multiplying previous
// term with (n - i + 1)/i
term = term * X * (n - i + 1)/(i * A);
document.write(term + " " );
}
} // main function started let A = 3, X = 4, n = 5;
series(A, X, n);
// This code is contributed by Surbhi Tyagi. </script> |
243 1620 4320 5760 3840 1024
Time complexity : O(n)
Auxiliary Space : O(1)