Question

If the number of integral terms in the expansion of \[{\left( {{3^{\dfrac{1}{2}}} + {5^{\dfrac{1}{8}}}} \right)^n}\] is exactly \[33\], then the least value of \[n\] is
A. \[128\]
B. \[248\]
C. \[256\]
D. \[264\]

Answer 1

Hint:
In this question, we have to find the least value of \[n\] for the expansion \[{\left( {{3^{\dfrac{1}{2}}} + {5^{\dfrac{1}{8}}}} \right)^n}\] First we will look at the number of expansion \[{\left( {a + b} \right)^n}\] and then with the help of that we will write the total number of terms and then we will find the L.C.M. of \[2\],\[8\] and then all the terms which has the power in multiple of \[8\] will be integer and then use the arithmetic progression formula to find the least value.

Formula used:
The basic expansion formula and arithmetic progression are given as
1. \[{\left( {a + b} \right)^n} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}\]
2. \[a + (n' - 1)d = {T_n}\]

Complete step-by-step solution:
Given that the expansion is\[{\left( {{3^{\dfrac{1}{2}}} + {5^{\dfrac{1}{8}}}} \right)^n}\]…(1)
Firstly, we will find the sequence of the given expansion using the formula \[{\left( {a + b} \right)^n} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}\].
From equation (1) \[a\] is \[{3^{\dfrac{1}{2}}}\] and \[b\] is \[{5^{\dfrac{1}{8}}}\].
Now, we will substitute the values in the formula, we get
\[{T_{r + 1}} = {}^n{C_r}{\left( {\sqrt 3 } \right)^{n - r}}{5^{\dfrac{r}{8}}}\]
For integral terms \[\dfrac{{n - r}}{2},\dfrac{r}{8}\] are both positive integers.
Here, \[n - r\] should be a multiple of \[2\].
Now, the LCM of \[2\] and \[8\] is \[8\].
Thus, we say that \[r\] should be divided by \[8\].
Now, the sequence we get is
\[r = 0,8,16,24,32,40,...\]
Here, \[r\] is less than or equal to \[n\] and greater than or equal to zero that is
\[0 \le r \le n\]
Further, we will find the least value of \[n\] using the arithmetic progression formula \[a + (n' - 1)d = {T_n}\]
Here, \[a\] is \[0,d\] is \[8\] , \[n'\] is \[33\] and \[{T_n} = n\]
Substitute the values in arithmetic progression formula, we get
\[\begin{array}{l}n = 0 + \left( {33 - 1} \right)8\\n = 32 \times 8\\n = 256\end{array}\]
Hence, the option (C) is correct

Note Another method to solve this question is to expand the given expression \[{\left( {{3^{\dfrac{1}{2}}} + {5^{\dfrac{1}{8}}}} \right)^n}\] by using the formula \[{\left( {a + b} \right)^n} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}\] and then counting all the integral terms or one can also try to find out the pattern that after how much terms we get the integral term and then use arithmetic progression to find the least value but it is time consuming.