Question

Find the number of terms in the expansion of ${{\left({{x}^{\dfrac{1}{3}}}+{{x}^{\dfrac{2}{5}}} \right)}^{40}}$ with integral power of

Answer 1

First of all, we are going to write the general form of the expression given in the above problem. The given expression is written in the form of ${{\left( x+y \right)}^{n}}$ and we know the general term in the expansion of this expression is equal to ${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}$. Now, substitute x as ${{x}^{\dfrac{1}{3}}}$ and y as ${{x}^{\dfrac{2}{5}}}$ and n as 40 given in the right hand side of the equation, we want integral power of x so rearrange the terms and see, on what values of r we are getting the integral power of x. Then count those terms which have integral power of x. These terms are the required answer.

Complete step-by-step answer:
The expression given in the above problem is as follows:
${{\left( {{x}^{\dfrac{1}{3}}}+{{x}^{\dfrac{2}{5}}} \right)}^{40}}$
The above expression is the binomial expression and the general term in the above binomial expansion is:
We know that if we have a binomial expansion ${{\left( x+y \right)}^{n}}$ then the general term in this expansion is:
${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}$
Now, substituting the value of x as ${{x}^{\dfrac{1}{3}}}$ and y as ${{x}^{\dfrac{2}{5}}}$ and n as 40 in the above equation we get,
$\begin{aligned} & {{T}_{r+1}}={}^{40}{{C}_{r}}{{x}^{\left( \dfrac{1}{3} \right)\left( 40-r \right)}}{{x}^{\left( \dfrac{2}{5} \right)r}} \\\ & \Rightarrow {{T}_{r+1}}={}^{40}{{C}_{r}}{{x}^{\dfrac{40-r}{3}+\left( \dfrac{2}{5} \right)r}} \\\ \end{aligned}$
It is asked in the question that we have to find the number of terms in which power of x is an integer so we are going to write power of x as follows:
$\dfrac{40-r}{3}+\dfrac{2}{5}r$
Taking 15 as L.C.M in the above expression we get,
$\begin{aligned} & \dfrac{5\left( 40-r \right)+2\left( 3 \right)r}{15} \\\ & =\dfrac{200-5r+6r}{15} \\\ & =\dfrac{200+r}{15} \\\ \end{aligned}$
Now, the above expression will give an integral answer when the value of r in such a way that addition with 200 will be divisible by 15.
When we put r as 10 then we get 210 which is divisible by 15. Similarly, the possible values of r so that
the number will be divisible by 15 are:
$10,25,40$
Hence, there are 3 terms in which the power of x is an integer.

Note: You might be thinking that why we have stopped at r equals 40 because the maximum value that r can take 40. In the expression given below:
${{T}_{r+1}}={}^{40}{{C}_{r}}{{x}^{\dfrac{40-r}{3}+\left( \dfrac{2}{5} \right)r}}$
We know that the maximum value that r can take is 40. That’s why we stopped at 40