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The 5th term of ${\left( {3a - 2b} \right)^{ - 1}}$

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Hint: Here we have to find the 5th term in the expansion of ${\left( {3a - 2b} \right)^{ - 1}}$, so use the formula for any general ${(r + 1)^{th}}$ term in the series expansion of ${(1 + x)^n}$.Note point is to simplify the given expression ${\left( {3a - 2b} \right)^{ - 1}}$into ${(1 + x)^n}$ before finding the general ${(r + 1)^{th}}$. This concept will help you reach the right answer to this problem statement.

We have been given the expression ${\left( {3a - 2b} \right)^{ - 1}}$ so let’s first simplify it to ${(1 + x)^n}$ form.

Taking common 3a from the expression ${\left( {3a - 2b} \right)^{ - 1}}$ we get,

$ \Rightarrow {\left( {3a} \right)^{ - 1}}{\left( {1 - \dfrac{{2b}}{{3a}}} \right)^{ - 1}} = \dfrac{1}{{3a}}{\left( {1 - \dfrac{{2b}}{{3a}}} \right)^{ - 1}}$……………………….. (1)

Any general ${(r + 1)^{th}}$ term in the expansion of ${(1 + x)^n}$ is given as ${T_{r + 1}} = \dfrac{{n(n - 1)(n - 2).............(n - r + 1)}}{{r!}}{x^r}$………………… (2)

We have to find 5th that is ${T_5}$ term so clearly r=4 (using equation 2)…………………… (3)

Now using equation (1) and on comparing with ${(1 + x)^n}$ we can say that $x = \dfrac{{ - 2b}}{{3a}}$ and $n = - 1$……………… (4)

Substituting the values from equation (4) and equation (3) in equation (2) and using equation (1) we get,

${T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{( - 1)( - 1 - 1)( - 1 - 2)( - 1 - 4 + 1)}}{{4!}}{{\left( {\dfrac{{ - 2b}}{{3a}}} \right)}^4}} \right]$

$ \Rightarrow {T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{( - 1)( - 2)( - 3)( - 4)}}{{4!}}{{\left( { - 1} \right)}^4}\left( {\dfrac{{16{b^4}}}{{81{a^4}}}} \right)} \right]$

On solving we get

$ \Rightarrow {T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{24}}{{4 \times 3 \times 2 \times 1}}\left( {\dfrac{{16{b^4}}}{{81{a^4}}}} \right)} \right]$

Thus the value of 5th term in the expansion of ${\left( {3a - 2b} \right)^{ - 1}}$is ${T_5} = \dfrac{{16{b^4}}}{{243{a^5}}}$.

Note: Whenever we face such type of problems the key concept to convert the given series expansion into the form ${(1 + x)^n}$ so that the general formula of any ${(r + 1)^{th}}$ term could be applicable to that given expression. This concept will help you get on the right track to get to the answer.

We have been given the expression ${\left( {3a - 2b} \right)^{ - 1}}$ so let’s first simplify it to ${(1 + x)^n}$ form.

Taking common 3a from the expression ${\left( {3a - 2b} \right)^{ - 1}}$ we get,

$ \Rightarrow {\left( {3a} \right)^{ - 1}}{\left( {1 - \dfrac{{2b}}{{3a}}} \right)^{ - 1}} = \dfrac{1}{{3a}}{\left( {1 - \dfrac{{2b}}{{3a}}} \right)^{ - 1}}$……………………….. (1)

Any general ${(r + 1)^{th}}$ term in the expansion of ${(1 + x)^n}$ is given as ${T_{r + 1}} = \dfrac{{n(n - 1)(n - 2).............(n - r + 1)}}{{r!}}{x^r}$………………… (2)

We have to find 5th that is ${T_5}$ term so clearly r=4 (using equation 2)…………………… (3)

Now using equation (1) and on comparing with ${(1 + x)^n}$ we can say that $x = \dfrac{{ - 2b}}{{3a}}$ and $n = - 1$……………… (4)

Substituting the values from equation (4) and equation (3) in equation (2) and using equation (1) we get,

${T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{( - 1)( - 1 - 1)( - 1 - 2)( - 1 - 4 + 1)}}{{4!}}{{\left( {\dfrac{{ - 2b}}{{3a}}} \right)}^4}} \right]$

$ \Rightarrow {T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{( - 1)( - 2)( - 3)( - 4)}}{{4!}}{{\left( { - 1} \right)}^4}\left( {\dfrac{{16{b^4}}}{{81{a^4}}}} \right)} \right]$

On solving we get

$ \Rightarrow {T_5} = \dfrac{1}{{3a}}\left[ {\dfrac{{24}}{{4 \times 3 \times 2 \times 1}}\left( {\dfrac{{16{b^4}}}{{81{a^4}}}} \right)} \right]$

Thus the value of 5th term in the expansion of ${\left( {3a - 2b} \right)^{ - 1}}$is ${T_5} = \dfrac{{16{b^4}}}{{243{a^5}}}$.

Note: Whenever we face such type of problems the key concept to convert the given series expansion into the form ${(1 + x)^n}$ so that the general formula of any ${(r + 1)^{th}}$ term could be applicable to that given expression. This concept will help you get on the right track to get to the answer.

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