Let n be a positive integer. Let A = Σ (k=0 to n) (-1)^k nCk [(1/2)^k + (3/4)^k + (7/8)^k + (15/16)^

Q: Wait, why can we just distribute the summation sign like that? Is that always allowed?

Yes, absolutely! The summation operator is linear. This means $\sum (a_k + b_k) = \sum a_k + \sum b_k$. Since our expression is a finite sum of terms, we can treat each bracketed term as a separate series, solve them individually, and then add them up. It is a powerful tool to break down complex problems into manageable pieces.

Q: I got stuck at the GP step. How do I know the common ratio is $(1/2)^n$?

Great observation! After simplifying the brackets, you get terms like $(1/2)^n, (1/4)^n, (1/8)^n$, and so on. If you write these as $(1/2)^n, ((1/2)^2)^n, ((1/2)^3)^n$, you can see that each term is the previous term multiplied by $(1/2)^n$. Thus, the common ratio $r$ is indeed $(1/2)^n$.

Q: Why did the $2^n$ terms cancel out in the fraction? I'm worried I missed a step.

Don't worry! When you have a fraction like $\frac{1/2^n}{(2^n-1)/2^n}$, the $2^n$ in the denominator of the numerator and the $2^n$ in the denominator of the denominator cancel out perfectly. It is just like dividing fractions: $\frac{a/b}{c/b} = \frac{a}{b} \cdot \frac{b}{c} = \frac{a}{c}$.

Q: Is it possible that $n$ could have two different values? I checked $2^n-1=63$ and $5n=30$.

In this specific problem, both conditions must be satisfied simultaneously. Since $n=6$ satisfies both $2^6-1=63$ and $5(6)=30$, we have a consistent solution. If they had yielded different values, we would have to re-evaluate our algebra, but here, the math is beautifully consistent!

Q: Do I really need to memorize the Binomial Theorem for this?

You don't need to memorize it as a chore, but rather see it as a pattern. The expression $\sum_{k=0}^n {}^nC_k x^k$ is the 'DNA' of binomial expansions. Whenever you see a summation involving ${}^nC_k$, your brain should immediately think of $(1+x)^n$. It is the key that unlocks the door to this entire problem.

Question

Let $n$ be a positive integer. Let $A = \sum_{k=0}^n (-1)^k {}^nC_k \left[ \left(\frac{1}{2}\right)^k + \left(rac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k \right]$. If $63 A = 1 - \frac{1}{2^{30}}$, then $n$ is equal to ____

Sigma Percentile · Accepted Answer

### The Elegance of Patterns: Unraveling the Binomial Mystery

Imagine you are standing before a complex, intimidating wall of numbers. At first glance, the expression

$$A = \sum_{k=0}^n (-1)^k {}^nC_k \left[ \left(\frac{1}{2}\right)^k + \left(\frac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k \right]$$

looks like a chaotic mess of fractions and powers. But in the world of JEE Advanced, chaos is often just order in disguise.

Our goal today is to peel back these layers and reveal the simple, elegant truth hidden underneath.

### Phase 1: The Power of Distribution

The first step in any great journey is to simplify your surroundings. We see a summation $\sum_{k=0}^n$ acting on a group of terms.

Because the summation operator is linear, we can distribute it across every single term inside the brackets. Think of it as inviting every term to the party individually.

When we distribute the $(-1)^k$ and the ${}^nC_k$ to each fraction, the expression transforms into:

$$A = \sum_{k=0}^n {}^nC_k \left(-\frac{1}{2}\right)^k + \sum_{k=0}^n {}^nC_k \left(-\frac{3}{4}\right)^k + \dots + \sum_{k=0}^n {}^nC_k \left(-\frac{31}{32}\right)^k$$

Suddenly, the "chaos" starts to look familiar. Does this not remind you of the fundamental **Binomial Theorem**?

Recall that $(1+x)^n = \sum_{k=0}^n {}^nC_k x^k$. By comparing our new summations to this identity, we can collapse each one into a simple binomial form.

### Phase 2: Collapsing the Summations

Let us look at the first term: $\sum_{k=0}^n {}^nC_k (-1/2)^k$. Here, $x = -1/2$.

Therefore, this entire sum is simply $(1 - 1/2)^n$. Applying this logic to all five terms, the expression for $A$ becomes remarkably clean:

$$A = \left(1 - \frac{1}{2}\right)^n + \left(1 - \frac{3}{4}\right)^n + \left(1 - \frac{7}{8}\right)^n + \left(1 - \frac{15}{16}\right)^n + \left(1 - \frac{31}{32}\right)^n$$

Wait—look at what happens inside those parentheses! $1 - 1/2 = 1/2$, $1 - 3/4 = 1/4$, and so on. We are left with:

$$A = \left(\frac{1}{2}\right)^n + \left(\frac{1}{4}\right)^n + \left(\frac{1}{8}\right)^n + \left(\frac{1}{16}\right)^n + \left(\frac{1}{32}\right)^n$$

### Phase 3: The Geometric Progression

We have arrived at a beautiful sequence. If we rewrite these terms using powers of $2$, we see:

$$A = \frac{1}{2^n} + \frac{1}{(2^2)^n} + \frac{1}{(2^3)^n} + \frac{1}{(2^4)^n} + \frac{1}{(2^5)^n}$$

This is a classic **Geometric Progression (GP)** with the first term $a = 1/2^n$ and a common ratio $r = 1/2^n$. Since there are $5$ terms, we use the sum formula $S_m = a \frac{1-r^m}{1-r}$:

$$A = \frac{1}{2^n} \left( \frac{1 - (1/2^n)^5}{1 - 1/2^n} \right)$$

After a bit of algebraic housekeeping—multiplying the numerator and denominator by $2^n$—we find the expression simplifies to:

$$A = \frac{1 - \frac{1}{2^{5n}}}{2^n - 1}$$

### Phase 4: The Final Comparison

We are almost there. The problem gives us the condition $63 A = 1 - \frac{1}{2^{30}}$.

If we rearrange our derived equation to $(2^n - 1) A = 1 - \frac{1}{2^{5n}}$, we can compare it directly to the given condition.

By matching the coefficients and the exponents, we get two simple equations: $2^n - 1 = 63$ and $5n = 30$. Both lead us to the same beautiful conclusion: **$n = 6$**.

Take a moment to appreciate this. What started as a terrifying summation ended as a simple comparison of powers. This is the beauty of mathematics—no matter how complex the problem appears, there is always a path to simplicity if you trust the fundamental principles.

Enter Numerical Value:

Visualized Solution

The Given Expression for $A$

The Binomial Expansion Tool

Distributing the Summation

Applying the Binomial Formula

Simplifying the Brackets

Identifying the Geometric Progression

Applying the GP Sum Formula

Simplifying the GP Sum

Rearranging for Comparison

Comparing the Equations

Solving for $n$

The Sigma Insight: Properties of Binomial Coefficients

The Elegance of Patterns

Phase 1

Phase 2

Phase 3

Phase 4

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