Question

Write the equation of the general terms in the binomial expansion of ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$.

Answer 1

We start solving the problem by recalling the definition of the general term in the binomial expansion ${{\left( a+b \right)}^{n}}$. We compare the given binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$ with ${{\left( a+b \right)}^{n}}$ to get the values of a and b. We then write the general form for the binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$ and we use the facts ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and ${{\left( ab \right)}^{m}}={{a}^{m}}.{{b}^{m}}$ to proceed through the problem. We then use ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and make necessary calculations required to get the desired result.

Complete step-by-step answer :
According to the problem, we have a binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$ and we need to find the general term of it.
We know that the general term of the binomial expansion ${{\left( a+b \right)}^{n}}$ is defined as ${}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, where $0\le r\le n$.
We have $a={{x}^{2}}$ and $b=-yx$ from the binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$.
So, we have the general term of the binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$ as ${}^{12}{{C}_{r}}\times {{\left( {{x}^{2}} \right)}^{12-r}}\times {{\left( -yx \right)}^{r}}$, $0\le r\le 12$.
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and ${{\left( ab \right)}^{m}}={{a}^{m}}.{{b}^{m}}$. We use these results in ${}^{12}{{C}_{r}}\times {{\left( {{x}^{2}} \right)}^{12-r}}\times {{\left( -yx \right)}^{r}}$.
So, ${}^{12}{{C}_{r}}\times {{x}^{2\times }}^{\left( 12-r \right)}\times {{\left( -1 \right)}^{r}}\times {{y}^{r}}\times {{x}^{r}}$, $0\le r\le 12$.
$\Rightarrow {{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-2r}}\times {{x}^{r}}\times {{y}^{r}}$, $0\le r\le 12$.
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We use this result in ${{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-2r}}\times {{x}^{r}}\times {{y}^{r}}$.
$\Rightarrow {{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-2r+r}}\times {{y}^{r}}$, $0\le r\le 12$.
$\Rightarrow {{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-r}}\times {{y}^{r}}$, $0\le r\le 12$.
So, we have found the general term of the binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$ as ${{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-r}}\times {{y}^{r}}$, $0\le r\le 12$.
∴ The general term of the binomial expansion ${{\left( {{x}^{2}}-yx \right)}^{12}}$, $x\ne 0$ is ${{\left( -1 \right)}^{r}}\times {}^{12}{{C}_{r}}\times {{x}^{24-r}}\times {{y}^{r}}$, $0\le r\le 12$.

Note : We need to make sure that the value of the terms inside the binomial expansion should not be equal to zero as calculating expansion for zero is meaningless. It will be easy to find the coefficient of any power of x by using the general term of the expansion. Similarly, we can expect problems to find the value of the middle term of the given binomial expansion.