Question

The first integral term in the expansion of ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ , is its
A. 2nd term
B. 3rd term
C. 4th term
D. 5th term

Answer 1

Hint : Use the rth term of a binomial expansion formula. For a term to be an integer, its power must be free from square root and cube roots. So expand ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ , and find which term is free from square roots and cube roots.
Binomial expansion of ${\left( {x + y} \right)^n}$ is ${}^n{C_0}{x^n}{y^0} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + ..... + {}^n{C_n}{x^0}{y^n}$
(r+1)th term of a binomial expansion is ${T_{r + 1}} = {}^n{C_r}{x^{n - r}}{y^r}$

Complete step-by-step answer :
We are given to find the first integral term in the expansion of ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$
On comparing ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ with ${\left( {x + y} \right)^n}$ , we get the value of x is $\sqrt 3 = {3^{\dfrac{1}{2}}}$ , the value of y is $\sqrt[3]{2} = {2^{\dfrac{1}{3}}}$ and the value of n is 9.
Binomial expansion of ${\left( {x + y} \right)^n}$ is ${}^n{C_0}{x^n}{y^0} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + ..... + {}^n{C_n}{x^0}{y^n}$ , where C represents a Combination.
Using binomial expansion, expand the given expression.
${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} \\\ \Rightarrow {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} = {}^9{C_0}{\left( {\sqrt 3 } \right)^9}{\left( {\sqrt[3]{2}} \right)^0} + {}^9{C_1}{\left( {\sqrt 3 } \right)^8}{\left( {\sqrt[3]{2}} \right)^1} + {}^9{C_2}{\left( {\sqrt 3 } \right)^7}{\left( {\sqrt[3]{2}} \right)^2} + {}^9{C_3}{\left( {\sqrt 3 } \right)^6}{\left( {\sqrt[3]{2}} \right)^3}.... + {}^9{C_9}{\left( {\sqrt 3 } \right)^0}{\left( {\sqrt[3]{2}} \right)^9} \\\ = 1.{\left( {{3^{\dfrac{1}{2}}}} \right)^9}.1 + 9.{\left( {{3^{\dfrac{1}{2}}}} \right)^8}.{\left( {{2^{\dfrac{1}{3}}}} \right)^1} + 36.{\left( {{3^{\dfrac{1}{2}}}} \right)^7}.{\left( {{2^{\dfrac{1}{3}}}} \right)^2} + 84.{\left( {{3^{\dfrac{1}{2}}}} \right)^6}.{\left( {{2^{\dfrac{1}{3}}}} \right)^3} + .... + 1.1.{\left( {{2^{\dfrac{1}{3}}}} \right)^9} \\\ = {3^{\dfrac{9}{2}}} + 9.\left( {{3^4}} \right).\left( {{2^{\dfrac{1}{3}}}} \right) + 36.\left( {{3^{\dfrac{7}{2}}}} \right).\left( {{2^{\dfrac{2}{3}}}} \right) + 84.\left( {{3^3}} \right).\left( {{2^1}} \right) + ... + {2^3} \\\$
As we can see in the above expansion, the first term has a fraction power, second term also has a fraction power, third term also has two fraction powers, but the fourth term has integer powers so the fourth term will be an integer.
Fourth term is $84 \times {3^3} \times 2 = 4536$
Therefore, the first integral term in the expansion of ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ is its fourth term.
So, the correct answer is “Option C”.

Note : Another approach
In a binomial expansion for a term to be an integer, it should have fractional powers; this means the term should not have square roots, cube roots or fourth roots.
So we need to find the first term of the expansion of ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ which is an integer.
(r+1)th term of a binomial expansion ${\left( {x + y} \right)^n}$ is ${T_{r + 1}} = {}^n{C_r}{x^{n - r}}{y^r}$
In the same way, (r+1)th term of ${\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9}$ is
${T_{r + 1}} = {}^9{C_r}{\left( {\sqrt 3 } \right)^{9 - r}}{\left( {\sqrt[3]{2}} \right)^r} \\\ \Rightarrow {T_{r + 1}} = {}^9{C_r}{\left( {{3^{\dfrac{1}{2}}}} \right)^{9 - r}}{\left( {{2^{\dfrac{1}{3}}}} \right)^r} \\\ \Rightarrow {T_{r + 1}} = {}^9{C_r}\left( {{3^{\dfrac{{9 - r}}{2}}}} \right)\left( {{2^{\dfrac{r}{3}}}} \right) \\\$
So here for the term to be integer, $\dfrac{{9 - r}}{2},\dfrac{r}{3}$ must also be integers.
So for these to be integers, $\left( {9 - r} \right)$ should be divisible by 2 and r should be divisible by 3
The least number which satisfies this condition is 3, So that 9-3=6 is divisible by 2 and 3 is divisible by 3.
So the first integral term when r is equal to 3 is ${T_{r + 1}} = {T_{3 + 1}} = {T_4}$ , fourth term.