If n is the number of irrational terms in the expansion of (3^1/4+5^1/8)^60, then (n-1) is…

Question

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

Accepted Answer

$${T_{r + 1}} = {}^{60}{C_r}{\left( {{3^{1/4}}} \right)^{60 - r}}{\left( {{5^{1/8}}} \right)^r}$$

rational if $${{60 - r} \over 4},{r \over 8}$$, both are whole numbers, $$r \in \{ 0,1,2,......60\} $$

$${{60 - r} \over 4} \in W \Rightarrow r \in \{ 0,4,8,....60\} $$

and $${r \over 8} \in W \Rightarrow r \in \{ 0,8,16,.....56\} $$

$$ \therefore $$ Common terms $$r \in \{ 0,8,16,.....56\} $$

So, 8 terms are rational

Then Irrational terms = $61 - 8 = 53 = n$

$$ \therefore $$ $$n - 1 = 52 = 13 \times {2^2}$$

Factors 1, 2, 4, 13, 26, 52

Solutions · 1