Question

The number of integral terms in \[{\left( {\sqrt 3 + \sqrt[8]{2}} \right)^{64}}\]is-
A.8
B.7
C.9
D.6

Answer 1

Hint: Here, we are required to find the number of integral terms in the given expression. We will find the general term of the given expression and then find the sequence of possible values which \[r\] can take, and finally we can find the total number of possible integral terms using arithmetic progressions.

Complete step-by-step answer:
The given expression is \[{\left( {\sqrt 3 + \sqrt[8]{2}} \right)^{64}}\].
Now, this can also be written as:
\[{\left[ {{{\left( 3 \right)}^{\dfrac{1}{2}}} + {{\left( 2 \right)}^{\dfrac{1}{8}}}} \right]^{64}}\]
Now, as we know, the general term of the expression\[{\left( {x + y} \right)^n}\]is \[{T_{r + 1}} = {}^n{C_r}.{\left( x \right)^{n - r}}{\left( y \right)^r}\].
Substituting the values \[x = {\left( 3 \right)^{\dfrac{1}{2}}}\], \[y = {\left( 2 \right)^{\dfrac{1}{8}}}\] and \[n = 64\] in \[{T_{r + 1}} = {}^n{C_r}.{\left( x \right)^{n - r}}{\left( y \right)^r}\], we can write the general term of the expression \[{\left[ {{{\left( 3 \right)}^{\dfrac{1}{2}}} + {{\left( 2 \right)}^{\dfrac{1}{8}}}} \right]^{64}}\] as:
\[{T_{r + 1}} = {}^{64}{C_r}.{\left( 3 \right)^{\dfrac{{64 - r}}{2}}}{\left( 2 \right)^{\dfrac{r}{8}}}\]
Now, if we see the exponential power of 3, we can say that:
\[0 \le r \le 64\]
And since, it should be divisible by 2, hence, its values should be:
\[r = 0,2,4,6,...64\]
Also, if we see the exponential power of 2, then \[r\]should be divisible by 8.
Hence, it can take values:
\[r = 0,8,16,24,...64\]
Therefore, it is forming as an Arithmetic Progression (A.P.)
Here, first term, \[a = 0\], common difference, \[d = 8\] and last term, \[{T_n} = 64\]
Substituting the values in the formula for last term \[{T_n} = a + \left( {n - 1} \right)d\], we get,
\[64 = 0 + \left( {n - 1} \right)8\]
\[ \Rightarrow 64 = \left( {n - 1} \right)8\]
Dividing both sides by 8, we get
\[ \Rightarrow 8 = \left( {n - 1} \right)\]
Adding 1 on both sides, we get
\[ \Rightarrow 8 + 1 = n\]
\[ \Rightarrow n = 9\]
Now, as we can see that \[r\] should be divisible by 2 as well as 8. Hence, when we would combine them together, since both are even numbers, we get that the number of integral terms in the expression\[{\left( {\sqrt 3 + \sqrt[8]{2}} \right)^{64}}\]is 9.
Therefore, option C is the correct answer.

Note: In this question we were required to find the number of integral terms in a given expression. We should know that ‘integral terms’ means such terms which are whole numbers and are not in the form of fractions. Hence, all the positive natural numbers such as 1,2,3,.. , 0 and all the negative numbers such as \[ - 1, - 1, - 3,...\] are integers. Hence, when we expanded the given expression and then found out that the number of ‘whole terms’ or ‘integral terms’ were 9.