Binomial Theorem Question - None

Question

In the binomial expansion of $(7^{1/5}-3^{1/10})^{60}$, find the number of irrational terms.

Accepted Answer

Step 1: General term in the expansion of $(7^{1/5}-3^{1/10})^{60}$ is:

$T_{r+1} = \binom{60}{r}(7^{1/5})^{60-r}(-3^{1/10})^r = \binom{60}{r}(-1)^r \cdot 7^{(60-r)/5} \cdot 3^{r/10}$

Step 2: For a term to be rational, both exponents must be integers:

• $(60-r)/5$ is an integer ⟹ $60-r \equiv 0 \pmod{5}$ ⟹ $r \equiv 0 \pmod{5}$

• $r/10$ is an integer ⟹ $r \equiv 0 \pmod{10}$

Step 3: Both conditions must hold simultaneously:

$r \equiv 0 \pmod{10}$ (since this is the stricter condition)

So $r \in \{0, 10, 20, 30, 40, 50, 60\}$ — that's 7 rational terms

Step 4: Total terms = 61 (from $r = 0$ to $r = 60$)

Number of irrational terms = $61 - 7 = 54$

∴ Answer: 54

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