If n is the number of irrational terms in the expansion of 31-4 - 51-8 60- then n-1 is divisible by-

T_r+1 = ^60C_r (3^1/4)^60 r (5^1/8)^r rational if 60 r4, r8 both are whole numbers, r ∈{0, 1, 2,....60} 60 r4∈ W ⇒ r ∈{0, 4, 8, ...60} and r8∈ W ⇒ r ∈{0, 8, 16, ...

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Byju's Answer

Standard XII

Mathematics

Number of Terms in Binomial Expansion

If n is the n...

Question

If $n$ is the number of irrational terms in the expansion of ${(3^{\frac{1}{4}} + 5^{\frac{1}{8}})}^{60}$ , then $(n - 1)$ is divisible by:

8

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26

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7

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30

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Solution

The correct option is B $26$
$T_{r + 1} =^{60} C_{r} (3^{1 / 4})^{60 - r} (5^{1 / 8})^{r}$
rational if $\frac{60 - r}{4}, \frac{r}{8}$ both are whole numbers, $r \in {0, 1, 2, . . . .60}$
$\frac{60 - r}{4} \in W \Rightarrow r \in {0, 4, 8, . . .60} and \frac{r}{8} \in W \Rightarrow r \in {0, 8, 16, . . .56}$
$∴$ Common terms $r \in {0, 8, 16, . . . . 56}$
So $8$ terms are rational
Then irrational terms $= 61 - 8 = 53 = n$
$∴ n - 1 = 52 = 13 \times 2^{2}$
Factors $1, 2, 4, 13, 26, 52$

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Match the elements of List I with List II

List I	List II
A) lf $λ$ be the number of terms which are integers, in the expansion of $(5^{\frac{1}{6}} + 7^{\frac{1}{9}})^{1824}$ , then $λ$ is divisible by	P) 2
B) lf $λ$ be the number of terms which are rational in the expansion of $(5^{\frac{1}{6}} + 2^{\frac{1}{8}})^{100}$ , then $λ$ is divisible by	Q) 3
C) lf $λ$ be the number of terms which are irrational in the expansion of $(3^{\frac{1}{4}} + 4^{\frac{1}{3}})^{99}$ , then $λ$ is divisible by	R) 7
	S) 13
	T) 17