JEE Main 2025 Apr 4 Shift 1, Mathematics Q12: Introduction to Binomial Expansion

JEE Main2025Apr 4, Shift 1Mathematics

In the expansion of $(32 + \frac{1}{3 3})^{n}$ , $n \in N$ , if the ratio of $1 5^{th}$ term from the beginning to the $1 5^{th}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ is:

A
4060
B
1040
C
2300
D
4960

Solution

General term: $T_{r + 1} = (r n) (2^{1/3})^{n - r} (3^{- 1/3})^{r}$ .

$1 5^{th}$ from beginning $\Rightarrow r = 14$ : $T_{15} = (14 n) 2^{(n - 14) /3} \cdot 3^{- 14/3}$ .

$1 5^{th}$ from the end is the $(n - 13)^{th}$ from the beginning $\Rightarrow r = n - 14$ : $T_{15}^{'} = (n - 14 n) 2^{14/3} \cdot 3^{- (n - 14) /3}$ .

Their ratio: $\frac{T _{15}}{T _{15}^{'}} = 2^{(n - 28) /3} \cdot 3^{(n - 28) /3}$ . After simplification we obtain $6^{(n - 28) /3} = \frac{1}{6} = 6^{- 1}$ , so $\frac{n - 28}{3} = - 1$ , giving $n = 25$ .

$(3 25) = \frac{25 \cdot 24 \cdot 23}{6} = 2300$ .

Concept behind this question

Introduction to Binomial ExpansionNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 4 Shift 1, Mathematics Q12: Introduction to Binomial Expansion

Concept behind this question

More previous year questions on Introduction to Binomial Expansion

Practice more Mathematics