The sum of the rational terms in the binomial expansion of (2^1/2+3^1/5)^10 is:

Question

The sum of the rational terms in the binomial expansion of $\left(2^{1/2}+3^{1/5}\right)^{10}$ is:

Accepted Answer

$$T_{r+1}=\binom{10}{r}\left(2^{1/2}\right)^{10-r}\left(3^{1/5}\right)^r$$

Rational when $\dfrac{10-r}2$ and $\dfrac r5$ are both integers, i.e. $r$ even and a multiple of 5: $r=0$ or $r=10$.

- $r=0$: $2^5=32$
- $r=10$: $3^2=9$

$$\text{Sum}=32+9=41$$

Solutions · 1