
Find the coefficient of ${x^7}$ in ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$
Answer
550.8k+ views
Hint:We have some formulas to find the expansion of the sum of two or more than two terms of power up to three, but to find the expansion of the sum of two terms with a power greater than three, we use the binomial theorem.
We have to find the coefficient of ${x^7}$ in ${(a{x^2} +
\dfrac{1}{{bx}})^{11}}$ , so we have to first expand ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ and then observe the coefficient of the term containing ${x^7}$ .
Complete step by step answer:
In the expansion of \[{(a + b)^n}\] , the ${(r + 1)^{th}}$ term is given as –
${T_{r + 1}}{ = ^{11}}{C_r}{a^{n - r}}{b^r}$
${(r + 1)^{th}}$ term of ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ is –
$
{T_{r + 1}}{ = ^{11}}{C_r}{(a{x^2})^{11 - r}}{(\dfrac{1}{{bx}})^r} \\
{T_{r + 1}}{ = ^{11}}{C_r}{a^{11 - r}}{x^{22 - 2r}}\dfrac{1}{{{b^r}{x^r}}} \\
\Rightarrow {T_{r + 1}}{ = ^{11}}{C_r}\dfrac{{{a^{11 - r}}}}{{{b^r}}}{x^{22 - 3r}} \\
$
Put $22 - 3r = 7$ to find the coefficient of ${x^7}$
$
\Rightarrow 22 - 7 = 3r \\
\Rightarrow r = 5 \\
$
Putting this value of r in the above equation, we get –
$
{T_{5 + !}}{ = ^{11}}{C_5}\dfrac{{{a^{11 - 5}}}}{{{b^5}}}{x^7} \\
\Rightarrow {T_6}{ = ^{11}}{C_5}\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
$
We know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , Using this In the above equation –
$
{T_6} = \dfrac{{11!}}{{5!\left( {11 - 5} \right)!}}\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
{T_6} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6!}}{{5 \times 4 \times 3 \times 2
\times 1 \times 6!}} \times \dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
{T_6} = 11 \times 2 \times 3 \times 7 \times \dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
\Rightarrow {T_6} = 432\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
$
Hence, the coefficient of ${x^7}$ in ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ is $432\dfrac{{{a^6}}}{{{b^5}}}$.
Note:The binomial theorem is used to describe the algebraic expansion in the powers of a binomial. There is a formula to find the expansion of that algebraic expression from which the expression for the ${r^{th}}$ term is derived. The power of an algebraic expression represents the number of times that expression is multiplied with itself.
For example, if the power of ${(a + b)^2}$ is two means that the term $(a + b)$ is multiplied by itself two times. In the given question, the power is 11 which means the given term is multiplied to itself 11 times.
Thus, the binomial theorem is a quick way to find the expansion of binomial expression. The term $^n{C_r}$ represents the combination and thus we use the formula of combination for its expansion.
We have to find the coefficient of ${x^7}$ in ${(a{x^2} +
\dfrac{1}{{bx}})^{11}}$ , so we have to first expand ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ and then observe the coefficient of the term containing ${x^7}$ .
Complete step by step answer:
In the expansion of \[{(a + b)^n}\] , the ${(r + 1)^{th}}$ term is given as –
${T_{r + 1}}{ = ^{11}}{C_r}{a^{n - r}}{b^r}$
${(r + 1)^{th}}$ term of ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ is –
$
{T_{r + 1}}{ = ^{11}}{C_r}{(a{x^2})^{11 - r}}{(\dfrac{1}{{bx}})^r} \\
{T_{r + 1}}{ = ^{11}}{C_r}{a^{11 - r}}{x^{22 - 2r}}\dfrac{1}{{{b^r}{x^r}}} \\
\Rightarrow {T_{r + 1}}{ = ^{11}}{C_r}\dfrac{{{a^{11 - r}}}}{{{b^r}}}{x^{22 - 3r}} \\
$
Put $22 - 3r = 7$ to find the coefficient of ${x^7}$
$
\Rightarrow 22 - 7 = 3r \\
\Rightarrow r = 5 \\
$
Putting this value of r in the above equation, we get –
$
{T_{5 + !}}{ = ^{11}}{C_5}\dfrac{{{a^{11 - 5}}}}{{{b^5}}}{x^7} \\
\Rightarrow {T_6}{ = ^{11}}{C_5}\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
$
We know that $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , Using this In the above equation –
$
{T_6} = \dfrac{{11!}}{{5!\left( {11 - 5} \right)!}}\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
{T_6} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6!}}{{5 \times 4 \times 3 \times 2
\times 1 \times 6!}} \times \dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
{T_6} = 11 \times 2 \times 3 \times 7 \times \dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
\Rightarrow {T_6} = 432\dfrac{{{a^6}}}{{{b^5}}}{x^7} \\
$
Hence, the coefficient of ${x^7}$ in ${(a{x^2} + \dfrac{1}{{bx}})^{11}}$ is $432\dfrac{{{a^6}}}{{{b^5}}}$.
Note:The binomial theorem is used to describe the algebraic expansion in the powers of a binomial. There is a formula to find the expansion of that algebraic expression from which the expression for the ${r^{th}}$ term is derived. The power of an algebraic expression represents the number of times that expression is multiplied with itself.
For example, if the power of ${(a + b)^2}$ is two means that the term $(a + b)$ is multiplied by itself two times. In the given question, the power is 11 which means the given term is multiplied to itself 11 times.
Thus, the binomial theorem is a quick way to find the expansion of binomial expression. The term $^n{C_r}$ represents the combination and thus we use the formula of combination for its expansion.
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