JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\frac{1}{n+1} {^nC_n} + \frac{1}{n} {^nC_{n-1}} + \dots + \frac{1}{2} {^nC_1} + {^nC_0} = \frac{1023}{10}$ , then $n$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Series ko Pehchanna

Binomial coefficients ki property $^nC_r = ^nC_{n-r}$ ka upyog karke, hum di gayi series ko is tarah likh sakte hain:

\frac{^nC_0}{n+1} + \frac{^nC_1}{n} + \dots + \frac{^nC_n}{1}

Yeh series darasal $(1+x)^n$ ke integration ka parinaam hai:

\int_{0}^{1} (1+x)^n dx = \frac{^nC_0}{1} + \frac{^nC_1}{2} + \dots + \frac{^nC_n}{n+1}

Step 2: Standard Formula ka Prayog

Is integral ka nishchit maan (definite value) hota hai:

\text{Sum} = \frac{2^{n+1} - 1}{n+1}

Step 3: Equation ko Solve Karna

Humein sawal mein sum $\frac{1023}{10}$ diya gaya hai, isliye:

\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10}

Step 4: Options se Verification

LHS mein $n=9$ rakhne par:

\text{LHS} = \frac{2^{9+1} - 1}{9+1}

\text{LHS} = \frac{2^{10} - 1}{10}

\text{LHS} = \frac{1024 - 1}{10} = \frac{1023}{10}

Kyunki $\text{LHS} = \text{RHS}$ , isliye $n = 9$ sahi hai.

Sahi Answer: (2) 9