JEE Main 2023 — Mathematics Algebra

Given expansion is ${({3}^{\frac{1}{2}}+{5}^{\frac{1}{4}})}^{680}$(321​+541​)680.

The general term in the expansion ${(x+a)}^{n}$(x+a)n is ${T}_{r+1}=Crn\cdot {x}^{n-r}\cdot {a}^{r}$Tr+1​=Crn⋅xn−r⋅ar

$$\Rightarrow {T}_{r+1}=Cr680{({3}^{\frac{1}{2}})}^{680-r}{({5}^{\frac{1}{4}})}^{r},0\leq r\leq 680$$⇒Tr+1​=Cr680(321​)680−r(541​)r,0≤r≤680

$$=Cr680{(3)}^{\frac{680-r}{2}}{(5)}^{\frac{r}{4}}$$=Cr680(3)2680−r​(5)4r​

For ${(3)}^{\frac{680-r}{2}}{(5)}^{\frac{r}{4}}$(3)2680−r​(5)4r​ to be rational $r$r should be a multiple of $4$4.

$$\Rightarrow r=0,4,8,12,.....680$$⇒r=0,4,8,12,.....680

$$\Rightarrow {a}_{n}=a+(n-1)d$$⇒an​=a+(n−1)d

$$\Rightarrow 680=0+(n-1)4$$⇒680=0+(n−1)4

$$\Rightarrow n=171$$⇒n=171

That means $r$r can take $171$171 values.

Hence, the required answer is $171$171.