Dados tres enteros A, X y n. La tarea es encontrar el término medio en la serie de expansión binomial.
Ejemplos:
Input : A = 1, X = 1, n = 6 Output : MiddleTerm = 20 Input : A = 2, X = 4, n = 7 Output : MiddleTerm1 = 35840, MiddleTerm2 = 71680
Acercarse
(A + X) n = n C 0 A n X 0 + n C 1 A n-1 X 1 + n C 2 A n-2 X 2 + ……… + n C n-1 A 1 X n-1 + n C n A 0 X n
El número total de términos en la expansión binomial de (A + X) n es (n + 1).
El término general en desarrollo binomial viene dado por:
T r+1 = nC r A n-r X r
Si n es un número par:
Sea m el término medio de la serie de expansión binomial, entonces
n = 2m
m = n / 2
Sabemos que habrá n + 1 término, entonces,
n + 1 = 2m + 1
En este caso, la voluntad sólo tiene un término medio. Este término medio es (m + 1) el término th .
Por lo tanto, el término medio
T m+1 = n C m A n-m X m
si n es un número impar:
Sea m el término medio de la serie de expansión binomial, luego
sea n = 2m + 1
m = (n-1) / 2
número de términos = n + 1 = 2m + 1 + 1 = 2m + 2
En este caso habrá dos términos medios. Estos términos medios serán (m + 1) th y (m + 2) th term.
Por lo tanto, los términos medios son:
T m+1 = n C (n-1)/2 A (n+1)/2 X (n-1)/2
T m+2 = n C (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 number
int 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 code
int 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 number
def 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 code
n = 5
A = 2
X = 3
# function call
findMiddleTerm(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 number
function 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 number
function 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>
Producción:
MiddleTerm1 = 720 MiddleTerm2 = 1080
Publicación traducida automáticamente
Artículo escrito por Manish_100 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA