The main objective of this question is to find the coefficient of the term $x^5y^8$ in the expansion of $(x+y)^{13}$ using the Binomial theorem or expansion.
The binomial theorem was first mentioned in the fourth century BC by Euclids, a famous Greek mathematician. The binomial theorem also known as binomial expansion in elementary algebra represents the algebraic expansion of binomial powers. The polynomial $(x + y)^n$ can be expanded into a sum incorporating terms of the type $ax^by^c$ in which the exponents $b$ and $c$ are non-negative integers with their sum being equal to $n$ and the coefficient $a$ of every term is a particular positive integer relying on $n$ and $b$. The value of the exponent in the expansion of the binomial theorem can be a fraction or a negative number. The analogous power expressions become one when an exponent is zero.
The binomial series identity $(x+y)^n=\sum\limits_{k=0}^{\infty}\dbinom{n}{k} x^ky^{n-k}$ is the most general form of the binomial theorem in which $\dbinom{n}{k}$ is a binomial coefficient and $n$ is a real number. The condition for the convergence of this series is; $n\geq0$, or $\left|\dfrac{x}{y}\right|<1$. The expansion of $(x+y)^n$ contains $(n+1)$ terms and the terms $x^n$ and $y^n$ are the first and last terms, respectively in the expansion.
Expert Answer
Using the binomial theorem for a positive integer $n$:
$(x+y)^n=\sum\limits_{k=0}^{n}\dbinom{n}{k} x^ky^{n-k}$
Since we have to find the coefficient of $x^5y^8$, so equating this term with $x^ky^{n-k}$ we get:
$k=5$ and $n-k=8$
Also, the comparison of $(x+y)^{13}$ with $(x+y)^n$ will give:
$n=13$
Now, to find the coefficient, we need to calculate $\dbinom{n}{k}=\dbinom{13}{5}$
Since $\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}$
So that, $\dbinom{13}{5}=\dfrac{13!}{5!(13-5)!}$
$=\dfrac{13!}{5!8!}$
$=\dfrac{13\cdot12\cdot11\cdot10\cdot9\cdot 8!}{5!8!}$
$=\dfrac{154440}{120}$
$=1287$
So, the coefficient of $x^5y^8$ is $1287$.
Example 1
Expand $(1+y)^4$ using the binomial series.
Solution
The binomial series is given by:
$(x+y)^n=\sum\limits_{k=0}^{n}\dbinom{n}{k} x^ky^{n-k}$
Here, $x=1$ and $n=4$ so:
$(1+y)^4=\sum\limits_{k=0}^{4}\dbinom{4}{k} x^ky^{4-k}$
Now, expand the series as:
$=\dbinom{4}{0} (1)^0y^{4-0}+\dbinom{4}{1} (1)^1y^{4-1}+\dbinom{4}{2} (1)^2y^{4-2}+\dbinom{4}{3} (1)^3y^{4-3}+\dbinom{4}{k} (1)^4y^{4-4}$
$=\dbinom{4}{0}y^4+\dbinom{4}{1}y^3+\dbinom{4}{2}y^2+\dbinom{4}{3}y+\dbinom{4}{4}$
$=\dfrac{4!}{0!(4-0)!}y^4+\dfrac{4!}{1!(4-1)!}y^3+\dfrac{4!}{2!(4-2)!}y^2+\dfrac{4!}{3!(4-3)!}y+\dfrac{4!}{4!(4-4)!}$
$(1+y)^4=y^4+4y^3+6y^2+4y+1$
Example 2
Find the $23\,rd$ term in the expansion of $(x+y)^{25}$.
Solution
The $k\,th$ term in the binomial expansion can expressed by the general formula:
$\dbinom{n}{k-1}x^{n-(k-1)}y^{k-1}$
Here, $n=25$ and $k=23$
So, the $23\,rd$ term can be found as:
$23 \,rd\, \text{term} =\dbinom{25}{23-1}x^{25-(23-1)}y^{23-1}$
$=\dbinom{25}{22}x^{25-23+1}y^{22}$
$=\dbinom{25}{22}x^{3}y^{22}$
$=\dfrac{25!}{22!(25-22)!}x^{3}y^{22}$
$=\dfrac{25!}{22!3!}x^{3}y^{22}$
$23 \,rd\, \text{term} =2300x^{3}y^{22}$
Example 3
Find the coefficient of $7\,th$ term in the expansion of $(x+2)^{10}$
Solution
The binomial series is given by:
$(x+y)^n=\sum\limits_{k=0}^{n}\dbinom{n}{k} x^ky^{n-k}$
Also, given that:
$y=2$, $n=10$ and $k=7$
First, find the $7\,th$ term as:
$7\,th \, \text{term} =\dbinom{10}{7-1}x^{10-(7-1)}y^{7-1}$
$=\dbinom{10}{6}x^{10-7+1}y^{6}$
$=\dbinom{10}{6}x^{4}y^{6}$
$=\dfrac{10!}{6!(10-6)!}x^{4}y^{6}$
$=\dfrac{10!}{6!4!}x^{4}y^{6}$
$7\,th \, \text{term}=210x^{4}y^{6}$
Hence the coefficient of $7\,th$ term is $210$.