Answer
Verified
480k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE