JEE Main 2025 (Online) 24th January Evening ShiftQuestion 1
Suppose $A$ and $B$ are the coefficients of $30^{\text{th}}$ and $12^{\text{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A=5B$, then $n$ is equal to:
Binomial Theorem · General Term and Middle Term
Solution
Concept. The $k$-th term of $(1+x)^N$ has coefficient ${}^{N}C_{k-1}$. Formulas used. (1) ${}^{N}C_r={}^{N}C_{N-r}$ and $\dfrac{{}^{N}C_{12}}{{}^{N}C_{11}}=\dfrac{N-11}{12}$Step 1 - Coefficients. With $N=2n-1$: $A={}^{N}C_{29}$, $B={}^{N}C_{11}$, and we need $\dfrac AB=\dfrac52$. Step 2 - The ratio grows with $N$. $$\frac{{}^{N}C_{29}}{{}^{N}C_{11}}=\frac{(N-11)(N-12)\cdots(N-28)}{12\cdot13\cdots29}$$ increases with $N$, so at most one $N$ works. Step 3 - Try $N=41$. By (1), ${}^{41}C_{29}={}^{41}C_{12}=\dfrac{30}{12}\,{}^{41}C_{11}=\dfrac52\,{}^{41}C_{11}$. So $N=41$ works. Step 4 - Value. $2n-1=41$ gives $n=21$. Answer: (D)