The number of terms in the expansion of (y^1/5+x^1/10)^55, in which powers of x and y are …

Question

The number of terms in the expansion of $\left(y^{1/5}+x^{1/10}\right)^{55}$, in which powers of $x$ and $y$ are free from radical signs, are:

Accepted Answer

$$T_{r+1}=\binom{55}{r}y^{(55-r)/5}x^{r/10}$$

Both $\dfrac{55-r}5$ and $\dfrac r{10}$ must be integers, so $r$ must be a multiple of 10, and $55-r$ a multiple of 5 (automatic). Valid $r=0,10,20,30,40,50$: six terms.

Solutions · 1