Question

The number of irrational terms in the expansion of ${(\sqrt[8]{5} + \sqrt[6]{2})^{100}}$ is
A.97
B.98
C.96
D.99

Answer 1

Hint: In this expansion, we must find the number of rational numbers and then subtract if from total numbers of terms to get the number of irrational terms that are obtained from the given expansion. For a number to be rational in an expansion of ${(\sqrt[m]{a} + \sqrt[n]{b})^z}$, the powers of a and b must be an integer.

Complete step-by-step answer:
The given expansion ${(\sqrt[8]{5} + \sqrt[6]{2})^{100}}$has a power of 100,
$ \Rightarrow $ Total number of terms in the expansion= 101
The ${r^{th}}$ term of the given expansion is given by,
${T_{r + 1}} = $ \[^n{C_r}{a^{n - r}}{b^r}\]
$ \Rightarrow $ ${T_{r + 1}} = $\[^{100}{C_r}{(\sqrt[8]{5})^{100 - r}}{(\sqrt[6]{2})^r}\]
$ \Rightarrow $${T_{r + 1}} = $\[^{100}{C_r}{({5^{\dfrac{1}{8}}})^{100 - r}}{({2^{\dfrac{1}{6}}})^r}\]
$ \Rightarrow $${T_{r + 1}} = $\[^{100}{C_r}{(5)^{\dfrac{{100 - r}}{8}}}{(2)^{\dfrac{r}{6}}}\]
Thus for a number to be rational, both the powers of 5 and 2 must be an integer.
Hence,
$\dfrac{{100 - r}}{8} = k$(Let k be an integer) and $\dfrac{r}{6} = l$(Let l be another integer)
$ \Rightarrow 100 - r = 8k$ And $r = 6k$
$ \Rightarrow 100 - r = 0,8,16,24,32,40,48,56,64,72,80,88,96$ And $r = 0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96$
$ \Rightarrow $The points where both terms of 5 and 2 can be rational are the common values between them i.e. 24,48,72,96.
Hence the total number of rational numbers in the expansion = 4
$ \Rightarrow $ Number of irrational numbers= 101-4
 $ \Rightarrow $Number of irrational numbers =97

Note: For a binomial expansion ${(a + b)^n}$ , the ${r^{th}}$ term is given by
${T_{r + 1}} = $ \[^n{C_r}{a^{n - r}}{b^r}\]
A rational number always has its power as an integer, always keep it in mind.