Σ (i=1 to 20) [20Ci-1 / (20Ci + 20Ci-1)] = k/21, then k equals : (1) 200 (2) 50 (3) 100 (4) 400

Q: Wait, why does ${}^{20}C_{i} + {}^{20}C_{i-1}$ become ${}^{21}C_{i}$? I thought Pascal's identity was only for specific numbers.

Pascal's Identity is a universal truth of combinatorics! It states that ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$. It represents the idea that choosing $r$ items from $n+1$ is the same as either including a specific item (choosing $r-1$ from $n$) or excluding it (choosing $r$ from $n$). It is the very heartbeat of Pascal's Triangle.

Q: Is there a shortcut to simplify $\frac{{}^{n}C_{r-1}}{{}^{n+1}C_{r}}$ without writing out all the factorials?

You can derive it quickly! Since ${}^{n+1}C_{r} = \frac{n+1}{r} \times {}^{n}C_{r-1}$, the ratio is simply $\frac{r}{n+1}$. Keeping this relationship in your toolkit saves precious seconds in the exam hall.

Q: Why did we cube the ratio? The original problem had the cube outside the fraction.

Exactly! The summation is over the cube of the simplified term. Once we reduced the fraction inside to $\frac{i}{21}$, the entire expression became $\sum (\frac{i}{21})^3$. Always keep an eye on the exponents—they are easy to miss in the heat of the moment!

Q: Can I solve this by expanding the binomial coefficients? It seems like a lot of work.

You could, but you would be trapped in a forest of factorials! The beauty of JEE problems lies in finding the elegant path. By using identities, we turn a nightmare of arithmetic into a clean, algebraic expression.

Q: How do I know when to stop simplifying and start summing?

Once your general term $T_i$ is a simple function of $i$ (like $i^3$ or $i^2$), you have reached the 'summation-ready' state. If you see a summation, your goal is always to isolate $i$ so you can use standard power sum formulas.

Question

$\sum_{i=1}^{20} \frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}} = \frac{k}{21}$, then k equals :

Options:

(A) 200

(B) 50

(C) 100

(D) 400

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a summation problem; we are peeling back the layers of a beautiful combinatorial structure.

When you first look at the expression

$$ \sum_{i=1}^{20} \left( \frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}} ight)^3 = \frac{k}{21} $$

it is natural to feel a bit overwhelmed. It looks like a mountain of factorials waiting to crush your confidence. But take a deep breath; in mathematics, the most complex-looking problems often hide a core of profound simplicity.

### The Pascal Revelation

Let us focus our gaze on the denominator: ${}^{20}C_i + {}^{20}C_{i-1}$. This is the classic signature of **Pascal's Identity**:

$$ {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} $$

This identity is the DNA of binomial coefficients. It tells us that the sum of two adjacent entries in one row of Pascal's triangle is simply the entry directly below them in the next row.

By applying this, our denominator transforms instantly from a sum into a single term: ${}^{21}C_i$. Suddenly, the expression becomes much cleaner:

$$ T_i = \frac{{}^{20}C_{i-1}}{{}^{21}C_i} $$

### The Ratio Simplification

Now, we have a ratio of binomial coefficients. Instead of expanding them into massive factorials, let us look for the relationship between them.

We know that

$$ {}^{n+1}C_{r} = \frac{n+1}{r} 	imes {}^{n}C_{r-1} $$

If we rearrange this, we find that

$$ \frac{{}^{n}C_{r-1}}{{}^{n+1}C_{r}} = \frac{r}{n+1} $$

Substituting $n=20$ and $r=i$, our term $T_i$ collapses beautifully into $\frac{i}{21}$. This is the power of pattern recognition in the JEE Advanced exam.

### The Sum of Cubes

With our simplified term $T_i = \frac{i}{21}$, the original summation becomes:

$$ \sum_{i=1}^{20} \left( \frac{i}{21} ight)^3 = \frac{k}{21} $$

We can pull the constant $\frac{1}{21^3}$ out of the summation, leaving us with:

$$ \frac{1}{21^3} \sum_{i=1}^{20} i^3 = \frac{k}{21} $$

Now, we invoke the classic identity for the sum of the first $n$ cubes:

$$ \sum_{i=1}^{n} i^3 = \left[ \frac{n(n+1)}{2} ight]^2 $$

For $n=20$, this is:

$$ \left[ \frac{20 	imes 21}{2} ight]^2 = (10 	imes 21)^2 = 100 	imes 21^2 $$

### Final Calculation

We are at the finish line. Substituting our sum back into the equation, we get:

$$ \frac{100 	imes 21^2}{21^3} = \frac{k}{21} $$

Notice how the powers of 21 cancel out with such grace. We are left with:

$$ \frac{100}{21} = \frac{k}{21} $$

Therefore, **$k = 100$**.

Look at what you have achieved. You didn't brute-force your way through; you used the structural beauty of binomial coefficients to dismantle the problem piece by piece. Keep this mindset—seek the symmetry, trust the identities, and never let the complexity of the notation intimidate you.

Animated Solution for Mathematics - Binomial Theorem: $\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :

Select Answer:

Visualized Solution

Analyze the General Term

Apply Pascal's Identity

Simplify the Ratio

Apply the Power of Three

Sum of Cubes Formula

Calculate the Sum Value

Equate and Solve for $k$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Pascal Revelation

The Ratio Simplification

The Sum of Cubes

Final Calculation

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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; } ∑i=120​20Ci​+20Ci−1​20Ci−1​​=21k​, then k equals :