JEE Main 2025 (Online) 8th April Evening ShiftQuestion 1
The number of integral terms in the expansion of $\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}$ is
Binomial Theorem · Rational and Integral Terms
Solution
Concept. A term is an integer when both powers of $5$ and $7$ are whole numbers. Formulas used. (1) General term of $(a+b)^n$: $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$(2) Number of terms of an A.P.: $\dfrac{\text{last}-\text{first}}{d}+1$Step 1 - General term. $$T_{r+1}={}^{1016}C_r\,5^{\frac{1016-r}{2}}\,7^{\frac r8}.$$ Step 2 - Conditions. We need $8\mid r$ and $2\mid(1016-r)$. If $8\mid r$, then $r$ is even, and $1016-r$ is even as well. So $8\mid r$ is enough. Step 3 - Count. $r=0,8,16,\dots,1016$. By (2): $\dfrac{1016-0}{8}+1=128$. Answer: (C)