Given three integers A, X, and n. The task is to find the middle term in the binomial expansion series.
Examples:
Input : A = 1, X = 1, n = 6
Output : MiddleTerm = 20Input : A = 2, X = 4, n = 7
Output : MiddleTerm1 = 35840, MiddleTerm2 = 71680
Approach
(A + X)n = nC0 An X0 + nC1 An-1 X1 + nC2 An-2 X2 + ……… + nCn-1 A1 Xn-1 + nCn A0 Xn
Total number of term in the binomial expansion of (A + X)n is (n + 1).
General term in binomial expansion is given by:
Tr+1 = nCr An-r Xr
If n is even number:
Let m be the middle term of binomial expansion series, then
n = 2m
m = n / 2
We know that there will be n + 1 term so,
n + 1 = 2m +1
In this case, there will is only one middle term. This middle term is (m + 1)th term.
Hence, the middle term
Tm+1 = nCmAn-mXm
if n is odd number:
Let m be the middle term of binomial expansion series, then
let n = 2m + 1
m = (n-1) / 2
number of terms = n + 1 = 2m + 1 + 1 = 2m + 2
In this case there will be two middle terms. These middle terms will be (m + 1)th and (m + 2)th term.
Hence, the middle terms are :
Tm+1 = nC(n-1)/2 A(n+1)/2 X(n-1)/2
Tm+2 = nC(n+1)/2 A(n-1)/2 X(n+1)/2
C++
// C++ program to find the middle term// in binomial expansion series. #include <bits/stdc++.h>using namespace std;// function to calculate // factorial of a numberint factorial(int n){ int fact = 1; for (int i = 1; i <= n; i++) fact *= i; return fact;}// Function to find middle term in // binomial expansion series.void findMiddleTerm(int A, int X, int n){ int i, j, aPow, xPow; float middleTerm1, middleTerm2; if (n % 2 == 0) { // If n is even // calculating the middle term i = n / 2; // calculating the value of A to // the power k and X to the power k aPow = (int)pow(A, n - i); xPow = (int)pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; cout << "MiddleTerm = " << middleTerm1 << endl; } else { // If n is odd // calculating the middle term i = (n - 1) / 2; j = (n + 1) / 2; // calculating the value of A to the // power k and X to the power k aPow = (int)pow(A, n - i); xPow = (int)pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; // calculating the value of A to the // power k and X to the power k aPow = (int)pow(A, n - j); xPow = (int)pow(X, j); middleTerm2 = ((float)factorial(n) / (factorial(n - j) * factorial(j))) * aPow * xPow; cout << "MiddleTerm1 = " << middleTerm1 << endl; cout << "MiddleTerm2 = " << middleTerm2 << endl; }}// Driver codeint main(){ int n = 5, A = 2, X = 3; // function call findMiddleTerm(A, X, n); return 0;} |
Java
// Java program to find the middle term// in binomial expansion series.import java.math.*;class GFG { // function to calculate factorial // of a number static int factorial(int n) { int fact = 1, i; if (n == 0) return 1; for (i = 1; i <= n; i++) fact *= i; return fact; } // Function to find middle term in // binomial expansion series. static void findmiddle(int A, int X, int n) { int i, j, aPow, xPow; float middleTerm1, middleTerm2; if (n % 2 == 0) { // If n is even // calculating the middle term i = n / 2; // calculating the value of A to // the power k and X to the power k aPow = (int)Math.pow(A, n - i); xPow = (int)Math.pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; System.out.println("MiddleTerm = " + middleTerm1); } else { // If n is odd // calculating the middle term i = (n - 1) / 2; j = (n + 1) / 2; // calculating the value of A to the // power k and X to the power k aPow = (int)Math.pow(A, n - i); xPow = (int)Math.pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; // calculating the value of A to the // power k and X to the power k aPow = (int)Math.pow(A, n - j); xPow = (int)Math.pow(X, j); middleTerm2 = ((float)factorial(n) / (factorial(n - j) * factorial(j))) * aPow * xPow; System.out.println("MiddleTerm1 = " + middleTerm1); System.out.println("MiddleTerm2 = " + middleTerm2); } } // Driver code public static void main(String[] args) { int n = 6, A = 2, X = 4; // calling the function findmiddle(A, X, n); }} |
Python3
# Python3 program to find the middle term# in binomial expansion series. import math# function to calculate # factorial of a numberdef factorial(n) : fact = 1 for i in range(1, n+1) : fact = fact * i return fact;# Function to find middle term in # binomial expansion series.def findMiddleTerm(A, X, n) : if (n % 2 == 0) : # If n is even # calculating the middle term i = int(n / 2) # calculating the value of A to # the power k and X to the power k aPow = int(math.pow(A, n - i)) xPow = int(math.pow(X, i)) middleTerm1 = ((math.factorial(n) / (math.factorial(n - i) * math.factorial(i))) * aPow * xPow) print ("MiddleTerm = {}" . format(middleTerm1)) else : # If n is odd # calculating the middle term i = int((n - 1) / 2) j = int((n + 1) / 2) # calculating the value of A to the # power k and X to the power k aPow = int(math.pow(A, n - i)) xPow = int(math.pow(X, i)) middleTerm1 = ((math.factorial(n) / (math.factorial(n - i) * math.factorial(i))) * aPow * xPow) # calculating the value of A to the # power k and X to the power k aPow = int(math.pow(A, n - j)) xPow = int(math.pow(X, j)) middleTerm2 = ((math.factorial(n) / (math.factorial(n - j) * math.factorial(j))) * aPow * xPow) print ("MiddleTerm1 = {}" . format(int(middleTerm1))) print ("MiddleTerm2 = {}" . format(int(middleTerm2)))# Driver coden = 5A = 2X = 3# function callfindMiddleTerm(A, X, n)# This code is contributed by # manishshaw1. |
C#
// C# program to find the middle term// in binomial expansion series.using System;class GFG { // function to calculate factorial // of a number static int factorial(int n) { int fact = 1, i; if (n == 0) return 1; for (i = 1; i <= n; i++) fact *= i; return fact; } // Function to find middle term in // binomial expansion series. static void findmiddle(int A, int X, int n) { int i, j, aPow, xPow; float middleTerm1, middleTerm2; if (n % 2 == 0) { // If n is even // calculating the middle term i = n / 2; // calculating the value of A to // the power k and X to the power k aPow = (int)Math.Pow(A, n - i); xPow = (int)Math.Pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; Console.WriteLine("MiddleTerm = " + middleTerm1); } else { // If n is odd // calculating the middle term i = (n - 1) / 2; j = (n + 1) / 2; // calculating the value of A to the // power k and X to the power k aPow = (int)Math.Pow(A, n - i); xPow = (int)Math.Pow(X, i); middleTerm1 = ((float)factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; // calculating the value of A to the // power k and X to the power k aPow = (int)Math.Pow(A, n - j); xPow = (int)Math.Pow(X, j); middleTerm2 = ((float)factorial(n) / (factorial(n - j) * factorial(j))) * aPow * xPow; Console.WriteLine("MiddleTerm1 = " + middleTerm1); Console.WriteLine("MiddleTerm2 = " + middleTerm2); } } // Driver code public static void Main() { int n = 5, A = 2, X = 3; // calling the function findmiddle(A, X, n); }}// This code is contributed by anuj_67. |
PHP
<?php// PHP program to find the middle term// in binomial expansion series. // function to calculate // factorial of a numberfunction factorial(int $n){ $fact = 1; for($i = 1; $i <= $n; $i++) $fact *= $i; return $fact;}// Function to find middle term in // binomial expansion series.function findMiddleTerm($A, $X, $n){ $i; $j; $aPow; $xPow; $middleTerm1; $middleTerm2; if ($n % 2 == 0) { // If n is even // calculating the middle term $i = $n / 2; // calculating the value of A to // the power k and X to the power k $aPow = pow($A, $n - i); $xPow = pow($X, $i); $middleTerm1 = $factorial($n) / (factorial($n - $i) * factorial($i)) * $aPow * $xPow; echo "MiddleTerm = ","\n", $middleTerm1 ; } else { // If n is odd // calculating the middle term $i = ($n - 1) / 2; $j = ($n + 1) / 2; // calculating the value of A to the // power k and X to the power k $aPow = pow($A, $n - $i); $xPow = pow($X, $i); $middleTerm1 = ((float)factorial($n) / (factorial($n - $i) * factorial($i))) * $aPow * $xPow; // calculating the value of A to the // power k and X to the power k $aPow = pow($A, $n - $j); $xPow = pow($X, $j); $middleTerm2 = factorial($n) / (factorial($n - $j) * factorial($j)) * $aPow * $xPow; echo "MiddleTerm1 = " , $middleTerm1,"\n" ; echo "MiddleTerm2 = " , $middleTerm2 ; }} // Driver code $n = 5; $A = 2; $X = 3; // function call findMiddleTerm($A, $X, $n);// This code is contributed by Vishal Tripathi.?> |
Javascript
<script>// JavaScript program to find the middle term// in binomial expansion series.// function to calculate// factorial of a numberfunction factorial(n){ let fact = 1; for (let i = 1; i <= n; i++) fact *= i; return fact;}// Function to find middle term in// binomial expansion series.function findMiddleTerm(A, X, n){ let i, j, aPow, xPow; let middleTerm1, middleTerm2; if (n % 2 == 0) { // If n is even // calculating the middle term i = Math.floor(n / 2); // calculating the value of A to // the power k and X to the power k aPow = Math.floor(Math.pow(A, n - i)); xPow = Math.floor(Math.pow(X, i)); middleTerm1 = Math.floor(factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; document.write("MiddleTerm = " + middleTerm1 + "<br>"); } else { // If n is odd // calculating the middle term i = Math.floor((n - 1) / 2); j = Math.floor((n + 1) / 2); // calculating the value of A to the // power k and X to the power k aPow = Math.floor(Math.pow(A, n - i)); xPow = Math.floor(Math.pow(X, i)); middleTerm1 = Math.floor(factorial(n) / (factorial(n - i) * factorial(i))) * aPow * xPow; // calculating the value of A to the // power k and X to the power k aPow = Math.floor(Math.pow(A, n - j)); xPow = Math.floor(Math.pow(X, j)); middleTerm2 = Math.floor(factorial(n) / (factorial(n - j) * factorial(j))) * aPow * xPow; document.write("MiddleTerm1 = " + middleTerm1 + "<br>"); document.write("MiddleTerm2 = " + middleTerm2 + "<br>"); }}// Driver code let n = 5, A = 2, X = 3; // function call findMiddleTerm(A, X, n);// This code is contributed by Surbhi Tyagi.</script> |
Output:
MiddleTerm1 = 720 MiddleTerm2 = 1080
Time Complexity: O(n)
Auxiliary Space: O(1)
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