Question

Find the natural number $m$ for which the coefficient of $x$ in the binomial expansion of $\left( {{x^m} + \dfrac{1}{{{x^2}}}} \right)$ is 1540.

Answer 1

Hint:For this question, we will first write the general term equation of a binomial expansion then we will equate the power of $x$ with 1 and the coefficient of $x$ with 1540. We do so because the coefficient of $x$ is given in the question. After this we find the value of $m$ such that it satisfies the condition of $r$in the general term equation which is: ${T_{r + 1}}{ = ^n}{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$ for the binomial expansion ${\left( {a + b} \right)^n}$ .

Formula Used:
${ }^{n} C_{r}=\dfrac{n !}{r ! .(n-r) !}$
The general term equation which is: ${T_{r + 1}}{ = ^n}{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$ for the binomial expansion ${\left( {a + b} \right)^n}$

Complete step by step Solution:
Given binomial expansion is $\left( {{x^m} + \dfrac{1}{{{x^2}}}} \right)$ .
The coefficient of $x$ is 1540.
Let us first write the general term formula of a binomial expansion ${\left( {a + b} \right)^n}$ as ${T_{r + 1}}{ = ^n}{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$ .
For the given binomial expansion this formula becomes: ${T_{r + 1}}{ = ^{22}}{C_r}{\left( {{x^m}} \right)^{22 - r}}{\left( {\dfrac{1}{{{x^2}}}} \right)^r}$
Simplifying it we get, ${T_{r + 1}}{ = ^{22}}{C_r}{x^{22m - mr - 2r}}$
Since we have been given the coefficient of $x$ , therefore we get the following two equations:
 $22m - mr - 2r = 1$ ...(1)
And $^{22}{C_r} = 1540$ ...(2)
Also we know that $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ ...(3)
Let us first find $r$ using equation numbers (2) and (3).
Here, we know that $r$would vary between 0 and 22 i.e., $0 \leqslant r \leqslant 22$ .
If we substitute different values of $r$ in equation number (2), we get that only $^{22}{C_3} = 1540$ and $^{22}{C_{19}} = 1540$ i.e., $r$ can either be 3 or 19.
 $r = 3\,or\,19$ ...(4)
Now we get $m$ in terms of $r$ using equation number (1).
 $22m - mr - 2r = 1$
This implies that $m(22 - r) - 2r = 1$
 $m = \dfrac{{1 + 2r}}{{22 - r}}$ ...(5)
Now we substitute both the values of $r$ from equation (4) into equation (5) one by one.
Now we know that $m$ and $r$ should be whole numbers so if this condition is not met then we do not consider the value of $m$ and $r$ respectively.
Let us first put $r = 3$ in equation (5).
We get that $m = \dfrac{{1 + 2 \times 3}}{{22 - 3}}$
Thus we find that $m = \dfrac{7}{{19}}$ which does not satisfy the condition mentioned above.
Therefore, $r \ne 3$ .
We do the same for $r = 19$ .
 $m = \dfrac{{1 + 2 \times 19}}{{22 - 19}}$
 We get $m = 13$ which satisfy our condition.
Hence, $m = 13$ is the correct answer.

Note: We do not always equate the power of $x$ with 1. We equate the power of $x$ according to the information given in the question. Also do not write the final answer without checking the condition mentioned in the solution. Similar questions can also be given for different types of expansion.