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Hint: First change the expression ${{\left( a+bx \right)}^{-1}}$ as $\dfrac{1}{a}{{\left( 1+\dfrac{b}{a}x \right)}^{-1}}$ and find its ${{\left( r+1 \right)}^{\text{th}}}$ term using the formula,

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$ to get what is asked in the question.

Complete step-by-step answer:

We have to find the ${{\left( r+1 \right)}^{\text{th}}}$ term of the expression \[{{\left( a+bx \right)}^{-1}}\].

We have to first write or mention the general term that is the ${{\left( r+1 \right)}^{\text{th}}}$ term of ${{\left( 1+x \right)}^{n}}$ which is given by the formula,

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}...........\left( i \right)$

Now let’s consider the expansion of ${{\left( a+bx \right)}^{-1}},$

Taking ‘a’ out of the bracket, we get

${{\left( a+bx \right)}^{-1}}={{\left\{ a\left( 1+\dfrac{b}{a}x \right) \right\}}^{-1}}$

Now we know the formula, ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ , so the above equation can be written as,

${{\left( a+bx \right)}^{-1}}={{a}^{-1}}{{\left( 1+\dfrac{b}{a}x \right)}^{-1}}$

So now it can be written as,

${{\left( a+bx \right)}^{-1}}=\dfrac{1}{a}{{\left( 1+\dfrac{b}{a}x \right)}^{-1}}$

Now we have to find ${{T}_{r+1}}$ of the expression $\dfrac{1}{a}{{\left( 1+\dfrac{b}{a} \right)}^{-1}}$ which is,

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right).........\left( -1-r+1 \right)}{r!} \right\}\times {{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}}$

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{\left( -1 \right)\left( -2 \right)\left( -3 \right)......\left( -r \right)}{r!}{{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}} \right\}$

In the expression (-1)(-2)(-3)…………(-r) can be written as ${{\left( -1 \right)}^{r}}.r!$, so the above equation becomes,

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{{{\left( -1 \right)}^{r}}r!}{r!}\times {{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}} \right\}$

Now combining the ‘a’ term, we get

${{T}_{r+1}}={{\left( -1 \right)}^{r}}\dfrac{{{b}^{r}}}{{{a}^{r+1}}}{{x}^{r}}$

Hence the ${{(r+1)}^{th}}$ term in the expansion of ${{\left( a+bx \right)}^{-1}}$ is ${{\left( -1 \right)}^{r}}\dfrac{{{b}^{r}}}{{{a}^{r+1}}}{{x}^{r}}.$

Note: Students must be careful while dealing with expansion related or identical to ${{\left( 1+x \right)}^{-n}}$ because in this the general formula of ${{T}_{r+1}}$ is expressed as,

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$

Irrespective of what value n is. One should also be careful about its calculation mistakes as the solution is too long.

Students generally make mistakes by applying the general formula of ${{T}_{r+1}}$ directly in${{\left( a+bx \right)}^{-1}}$ . They will get wrong answer as ${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$ is the general formula for rth in expansion of ${{\left( 1+x \right)}^{-n}}$, that means one of the term should be 1.

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$ to get what is asked in the question.

Complete step-by-step answer:

We have to find the ${{\left( r+1 \right)}^{\text{th}}}$ term of the expression \[{{\left( a+bx \right)}^{-1}}\].

We have to first write or mention the general term that is the ${{\left( r+1 \right)}^{\text{th}}}$ term of ${{\left( 1+x \right)}^{n}}$ which is given by the formula,

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}...........\left( i \right)$

Now let’s consider the expansion of ${{\left( a+bx \right)}^{-1}},$

Taking ‘a’ out of the bracket, we get

${{\left( a+bx \right)}^{-1}}={{\left\{ a\left( 1+\dfrac{b}{a}x \right) \right\}}^{-1}}$

Now we know the formula, ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ , so the above equation can be written as,

${{\left( a+bx \right)}^{-1}}={{a}^{-1}}{{\left( 1+\dfrac{b}{a}x \right)}^{-1}}$

So now it can be written as,

${{\left( a+bx \right)}^{-1}}=\dfrac{1}{a}{{\left( 1+\dfrac{b}{a}x \right)}^{-1}}$

Now we have to find ${{T}_{r+1}}$ of the expression $\dfrac{1}{a}{{\left( 1+\dfrac{b}{a} \right)}^{-1}}$ which is,

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right).........\left( -1-r+1 \right)}{r!} \right\}\times {{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}}$

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{\left( -1 \right)\left( -2 \right)\left( -3 \right)......\left( -r \right)}{r!}{{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}} \right\}$

In the expression (-1)(-2)(-3)…………(-r) can be written as ${{\left( -1 \right)}^{r}}.r!$, so the above equation becomes,

${{T}_{r+1}}=\dfrac{1}{a}\left\{ \dfrac{{{\left( -1 \right)}^{r}}r!}{r!}\times {{\left( \dfrac{b}{a} \right)}^{r}}\times {{x}^{r}} \right\}$

Now combining the ‘a’ term, we get

${{T}_{r+1}}={{\left( -1 \right)}^{r}}\dfrac{{{b}^{r}}}{{{a}^{r+1}}}{{x}^{r}}$

Hence the ${{(r+1)}^{th}}$ term in the expansion of ${{\left( a+bx \right)}^{-1}}$ is ${{\left( -1 \right)}^{r}}\dfrac{{{b}^{r}}}{{{a}^{r+1}}}{{x}^{r}}.$

Note: Students must be careful while dealing with expansion related or identical to ${{\left( 1+x \right)}^{-n}}$ because in this the general formula of ${{T}_{r+1}}$ is expressed as,

${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$

Irrespective of what value n is. One should also be careful about its calculation mistakes as the solution is too long.

Students generally make mistakes by applying the general formula of ${{T}_{r+1}}$ directly in${{\left( a+bx \right)}^{-1}}$ . They will get wrong answer as ${{T}_{r+1}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)..........\left( n-r+1 \right)}{r!}{{x}^{r}}$ is the general formula for rth in expansion of ${{\left( 1+x \right)}^{-n}}$, that means one of the term should be 1.

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