If 1/(n+1) ⁿCₙ + 1/n ⁿCₙ₋₁ + ... + 1/2 ⁿC₁ + ⁿC₀ = 1023/10 then n is equal to

Q: Why can't I just expand the binomial coefficients and add them up?

Expanding is a trap! In JEE Advanced, we look for patterns and properties. The absorption identity is designed to simplify exactly this type of expression. If you try to expand, you will be stuck in a sea of factorials that will take too long to solve.

Q: Where did the ${}^{n+1}C_0$ go in the summation?

Great observation! The summation index $r$ goes from $0$ to $n$, so the binomial coefficient index $r+1$ goes from $1$ to $n+1$. The term ${}^{n+1}C_0$ is indeed missing from the standard $2^{n+1}$ sum. We must subtract it to maintain equality.

Q: Is it okay to just guess $n=9$ by looking at the options?

In an exam, checking options is a valid strategy. However, understanding the derivation ensures you can solve variations of this problem where the answer isn't an integer or isn't obvious. Always prioritize the derivation first!

Question

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

Options:

(A) 9

(B) 8

(C) 7

(D) 6

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Patterns

Welcome, future engineer. Today, we are going to peel back the layers of a problem that looks intimidating but is actually a masterpiece of algebraic elegance.

When you first look at the series 
$$S = \frac{1}{n+1} {}^nC_n + \frac{1}{n} {}^nC_{n-1} + \dots + \frac{1}{2} {}^nC_1 + {}^nC_0$$
your instinct might be to panic. You see fractions, binomial coefficients, and a variable $n$ hiding in plain sight. But in the world of JEE Advanced, complexity is often just a mask for a beautiful, hidden symmetry.

### Phase 1: The Anatomy of the Series

Let us look at the general term. If we write this series using summation notation, we can define the $r$-th term as $T_r = \frac{1}{r+1} {}^nC_r$.

As $r$ ranges from $0$ to $n$, we generate every single term in that series. This is our first victory. We have transformed a messy, sprawling expression into a compact, mathematical object: 
$$S = \sum_{r=0}^n \frac{1}{r+1} {}^nC_r$$

### Phase 2: The Magic of Absorption

Here is where the magic happens. We encounter the term $\frac{1}{r+1} {}^nC_r$. If you try to expand this using factorials, you will find yourself in a labyrinth of algebra.

Instead, we invoke the **Binomial Absorption Identity**. This identity is a powerful tool in your arsenal: 
$$\frac{1}{r+1} {}^nC_r = \frac{1}{n+1} {}^{n+1}C_{r+1}$$
Think of it as a way to 'absorb' the fraction into the binomial coefficient, effectively shifting the indices. It is elegant, it is clean, and it is exactly what we need.

### Phase 3: The Summation Dance

Now, we substitute this identity back into our sum. We get:
$$S = \sum_{r=0}^n \frac{1}{n+1} {}^{n+1}C_{r+1}$$

Since $\frac{1}{n+1}$ is a constant with respect to $r$, we pull it out: 
$$S = \frac{1}{n+1} \sum_{r=0}^n {}^{n+1}C_{r+1}$$

Now, look closely at the summation. As $r$ goes from $0$ to $n$, the index $r+1$ goes from $1$ to $n+1$. This means we are summing ${}^{n+1}C_1 + {}^{n+1}C_2 + \dots + {}^{n+1}C_{n+1}$.

We know that the sum of all binomial coefficients $\sum_{k=0}^{N} {}^NC_k = 2^N$. Here, $N = n+1$. So, the full sum would be $2^{n+1}$. But our sum is missing the very first term, ${}^{n+1}C_0$. Since ${}^{n+1}C_0 = 1$, our sum is simply $2^{n+1} - 1$.

### Phase 4: The Final Reveal

We are almost there. Substituting this back, we have:
$$S = \frac{2^{n+1} - 1}{n+1}$$

The problem tells us that $S = \frac{1023}{10}$. So, we equate: 
$$\frac{2^{n+1} - 1}{n+1} = \frac{1023}{10}$$

By comparing the denominators, we immediately see that $n+1 = 10$, which means **$n = 9$**.

Let us verify: $2^{9+1} - 1 = 2^{10} - 1 = 1024 - 1 = 1023$. It matches perfectly! You have just navigated a complex series problem with precision and logic. Remember, the math didn't change; your perspective did.

Animated Solution for Mathematics - Binomial Theorem: If $\frac{1}{n + 1}^{n} C_{n} + \frac{1}{n}^{n} C_{n - 1} + \dots + \frac{1}{2}^{n} C_{1} +^{n} C_{0} = \frac{1023}{10}$ then $n$ is equal to

Select Answer:

Visualized Solution

Analyze the Given Series

Identify the General Term

The Binomial Absorption Identity

Rewrite the Summation

Expand the New Summation

Apply Sum of Coefficients Property

Form the Final Equation

Solve for n by Comparison

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Patterns

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n+11​nCn​+n1​nCn−1​+⋯+21​nC1​+nC0​=101023​ then n is equal to