JEE Main 2025 — 8 April, Evening Shift — Question 39

General Term Formula The general term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Substituting the given values $n=1016$, $a=5^{1/2}$, and $b=7^{1/8}$:

$$T_{r+1} = \binom{1016}{r} \left(5^{\frac{1}{2}}\right)^{1016-r} \left(7^{\frac{1}{8}}\right)^r$$ $$T_{r+1} = \binom{1016}{r} \cdot 5^{\frac{1016-r}{2}} \cdot 7^{\frac{r}{8}}$$

where $r \in \{0, 1, 2, \dots, 1016\}$.

Conditions for Integrality For the term to be an integer, the powers of the prime bases (5 and 7) must be non-negative integers. This leads to the following conditions: $\frac{1016-r}{2}$ must be an integer $\implies (1016-r)$ must be even $\implies r$ must be even. $\frac{r}{8}$ must be an integer $\implies r$ must be a multiple of $8$.

Since any multiple of 8 is automatically even, the governing condition is: $r$must be a multiple of $8$

Calculation We need to find the number of values of $r$ in the set $\{0, 1, 2, \dots, 1016\}$ that are divisible by 8. These values form an Arithmetic Progression (A.P.):

$$0, 8, 16, \dots, 1016$$

Using the formula for the $n^{th}$ term of an A.P., $l = a + (n-1)d$: First term ($a$) = 0 Common difference ($d$) = 8 Last term ($l$) = 1016

$$1016 = 0 + (n-1)8$$ $$\frac{1016}{8} = n - 1$$ $$127 = n - 1$$ $$n = 128$$

The number of integral terms is $128.$