JEE Mathematics 2025 Question with Solution

Given: We need the sum of all rational terms in $\left( 1 + 2^{1/3} + 3^{1/2} \right)^6$.

Find: The numerical value of the sum of all rational terms.

Using the multinomial theorem, a general term is

$$\frac{6!}{r_1!r_2!r_3!}\,(1)^{r_1}\,(2^{1/3})^{r_2}\,(3^{1/2})^{r_3}$$

where

$$r_1+r_2+r_3=6.$$

For the term to be rational, the exponents of $2$ and $3$ must be integers. Therefore:

$$r_2 \text{ must be divisible by } 3, \qquad r_3 \text{ must be even.}$$

Now list all triples $\left(r_1,r_2,r_3\right)$ satisfying these conditions and $r_1+r_2+r_3=6$:

1. $\left(6,0,0\right)$ gives value $1$.
2. $\left(4,0,2\right)$ gives coefficient $\dfrac{6!}{4!0!2!}=15$, so term value $15\cdot 3=45$.
3. $\left(2,0,4\right)$ gives coefficient $\dfrac{6!}{2!0!4!}=15$, so term value $15\cdot 3^2=135$.
4. $\left(0,0,6\right)$ gives value $3^3=27$.
5. $\left(3,3,0\right)$ gives coefficient $\dfrac{6!}{3!3!0!}=20$, so term value $20\cdot 2=40$.
6. $\left(1,3,2\right)$ gives coefficient $\dfrac{6!}{1!3!2!}=60$, so term value $60\cdot 2\cdot 3=360$.
7. $\left(0,6,0\right)$ gives value $2^2=4$.

Hence the sum of all rational terms is

$$1+45+135+27+40+360+4=612.$$

Therefore, the required sum is $612$.

Given: $\left(1+2^{1/3}+3^{1/2}\right)^6$

Find: The sum of all rational terms.

A multinomial term is

$$T=\frac{6!}{r_1!r_2!r_3!}\,1^{r_1}(2^{1/3})^{r_2}(3^{1/2})^{r_3}, \qquad r_1+r_2+r_3=6.$$

This becomes

$$T=\frac{6!}{r_1!r_2!r_3!}\,2^{r_2/3}3^{r_3/2}.$$

For $T$ to be rational, both $\dfrac{r_2}{3}$ and $\dfrac{r_3}{2}$ must be integers. So:

• $r_2 \in \{0,3,6\}$
• $r_3 \in \{0,2,4,6\}$

Now use $r_1=6-r_2-r_3 \ge 0$.

If $r_2=0$:

• $r_3=0 \Rightarrow (r_1,r_2,r_3)=(6,0,0)$, term $=1$
• $r_3=2 \Rightarrow (4,0,2)$, term $=\dfrac{6!}{4!2!}\cdot 3=15\cdot 3=45$
• $r_3=4 \Rightarrow (2,0,4)$, term $=\dfrac{6!}{2!4!}\cdot 3^2=15\cdot 9=135$
• $r_3=6 \Rightarrow (0,0,6)$, term $=3^3=27$

If $r_2=3$:

• $r_3=0 \Rightarrow (3,3,0)$, term $=\dfrac{6!}{3!3!}\cdot 2=20\cdot 2=40$
• $r_3=2 \Rightarrow (1,3,2)$, term $=\dfrac{6!}{1!3!2!}\cdot 2\cdot 3=60\cdot 6=360$
• $r_3=4$ is not possible because then $r_1<0$

If $r_2=6$:

• $r_3=0 \Rightarrow (0,6,0)$, term $=2^2=4$

Adding all rational terms:

$$1+45+135+27+40+360+4=612.$$

Therefore, the sum of all rational terms is $612$.