Question

If $n$ is the number of irrational terms in the expansion of ${\left[ {{3^{\dfrac{1}{4}}} + {5^{\dfrac{1}{8}}}} \right]^{60}}$ , then $(n - 1)$ is divisible by
a. $8$
b. $26$
c. $7$
d. $30$

Answer 1

Hint: Apply binomial formula as the given term is in the form of ${(a + b)^n}$i.e., ${\left[ {{3^{\dfrac{1}{4}}} + {5^{\dfrac{1}{8}}}} \right]^{60}}$where $n = 60$ is a positive integer. Then first find the rational terms by taking the power of $b$ as the other variable and also to know the total number of irrational terms and find the difference of total numbers and rational numbers.

Formula Used:
Binomial theorem formula –
${(a + b)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} $
Here, $a$ and $b$ are real numbers and $n$ is a positive integer where \[0 < r \leqslant n\].

Complete step by step solution: 
Given that,
${\left[ {{3^{\dfrac{1}{4}}} + {5^{\dfrac{1}{8}}}} \right]^{60}}$
Using Binomial theorem formula for the expansion of ${\left[ {{3^{\dfrac{1}{4}}} + {5^{\dfrac{1}{8}}}} \right]^{60}}$,
$ = {}^{60}{C_r}{\left( {{3^{\dfrac{1}{4}}}} \right)^{60 - r}}{\left( {{5^{\dfrac{1}{8}}}} \right)^r}$
$ = {}^{60}{C_r}{\left( 3 \right)^{\dfrac{{60 - r}}{4}}}{\left( 5 \right)^{\dfrac{r}{8}}},0 \leqslant r \leqslant 60$
Let, $\dfrac{r}{8} = p$
$ \Rightarrow r = 8p$
As we know that,
$0 \leqslant r \leqslant 60$
$0 \leqslant 8p \leqslant 60$
Divide all sides by $8$,
$0 \leqslant p \leqslant \dfrac{{60}}{8}$
$0 \leqslant p \leqslant 7.5$
$ \Rightarrow p = 0,1,2,3,4,5,6,7$
Since $\dfrac{{60 - r}}{4}$is divisible by $4$for all values of $p$,
Therefore, Total rational terms are Eight i.e., $0,1,2,3,4,5,6,7$
Now, \[Irrational{\text{ }}Terms = Total{\text{ }}Terms - Rational{\text{ }}Terms\]
\[Irrational{\text{ }}Terms = 61 - 8\]
\[Irrational{\text{ }}Terms = 53\]
Fore $n - 1 = 53 - 1 = 52$
Where, $n$ is the number of irrational terms (given)
$ \Rightarrow 52$ is divisible by $26$
Hence, the correct option is (b) i.e., $26$.

Note: Apply binomial expansion formula $\left( {{{(a + b)}^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} } \right)$ only when $n$ will be a positive integer. Also, while counting the total terms don’t forget to count $0$ with all the numbers. The procedure for calculating a binomial with an irrational exponent is identical to that for solving a binomial with a rational exponent.