Detailed Solution

1. Key Concepts and Formulas Used

This problem requires the application of two fundamental concepts from Combinatorics and Series:

• Pascal's Identity (or Addition Formula for Binomial Coefficients): This identity simplifies the sum of two adjacent binomial coefficients. It states that for non-negative integers $n$ and $r$ with $1 \le r \le n$, ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$ This identity is crucial for simplifying the denominator of the given expression.

• Sum of the First $n$ Cubes: The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum\limits_{i = 1}^{n} {{i^3}} = {\left( {{{n(n+1)} \over 2}} \right)^2}$ This formula will be used to evaluate the resulting summation.

2. Step-by-Step Working

We are asked to find the value of $k$ given the equation: ${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$

Step 2.1: Simplify the Denominator using Pascal's Identity

Why this step? The denominator ${}^{20}{C_i} + {}^{20}{C_{i-1}}$ is a sum of two binomial coefficients. Recognizing this structure allows us to apply Pascal's Identity to simplify it into a single binomial coefficient, making the expression much easier to handle.

Applying Pascal's Identity ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$ With $n=20$ and $r=i$, we have: ${}^{20}{C_i} + {}^{20}{C_{i-1}} = {}^{20+1}{C_i} = {}^{21}{C_i}$ So, the term inside the summation becomes: ${{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}$

Step 2.2: Simplify the Binomial Coefficient Ratio

Why this step? We now have a ratio of two binomial coefficients. Direct expansion using the definition ${}^{n}{C_r} = {n! \over {r!(n-r)!}}$ and subsequent simplification will reduce this complex ratio to a simple fractional form involving $i$. This is a standard technique for simplifying such ratios.

Let's expand the binomial coefficients: ${{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}} = {{{20!} \over {(i-1)!(20-(i-1))!}} \over {{{21!} \over {i!(21-i)!}}}}$ $= {{{20!} \over {(i-1)!(21-i)!}} \times {{i!(21-i)!} \over {21!}}}$ Now, we can cancel out common terms and expand factorials: $= {{{20!} \times i! \times (21-i)!} \over {(i-1)! \times (21-i)! \times 21!}}$ $= {{{20!} \times i \times (i-1)!} \over {(i-1)! \times 21 \times 20!}}$ Cancelling $(i-1)!$ and $20!$: $= {i \over {21}}$

Step 2.3: Substitute and Reformulate the Sum

Why this step? Having simplified the general term of the summation, we substitute it back into the original equation. This transforms the complex sum of binomial coefficient ratios into a much simpler sum of algebraic terms, allowing us to use standard summation formulas.

Substitute ${i \over {21}}$ back into the original sum: ${\sum\limits_{i = 1}^{20} {\left( {{i \over {21}}} \right)} ^3} = {k \over {21}}$ ${\sum\limits_{i = 1}^{20} {{{i^3}} \over {{{21}^3}}} } = {k \over {21}}$ We can factor out the constant term ${1 \over {{{21}^3}}}$ from the summation: ${1 \over {{{21}^3}}} \sum\limits_{i = 1}^{20} {{i^3}} = {k \over {21}}$

Step 2.4: Evaluate the Sum of Cubes

Why this step? We now need to calculate the sum of the first 20 cubes. Using the known formula for the sum of cubes is the most efficient way to do this.

Using the formula for the sum of the first $n$ cubes, $\sum\limits_{i = 1}^{n} {{i^3}} = {\left( {{{n(n+1)} \over 2}} \right)^2}$, with $n=20$: $\sum\limits_{i = 1}^{20} {{i^3}} = {\left( {{{20(20+1)} \over 2}} \right)^2}$ $= {\left( {{{20 \times 21} \over 2}} \right)^2}$ $= {\left( {10 \times 21} \right)^2}$ $= {210^2}$

Step 2.5: Solve for $k$

Why this step? With the sum evaluated, we substitute its value back into the equation from Step 2.3 and perform simple algebraic manipulation to isolate and find the value of $k$.

Substitute the value of $\sum\limits_{i = 1}^{20} {{i^3}} = {210^2}$ back into the equation: ${1 \over {{{21}^3}}} \times {210^2} = {k \over {21}}$ Recall that $210 = 10 \times 21$. So, $210^2 = (10 \times 21)^2 = 10^2 \times 21^2$. ${1 \over {{{21}^3}}} \times ({10^2} \times {21^2}) = {k \over {21}}$ $= {{{10^2} \times {21^2}} \over {{{21}^3}}} = {k \over {21}}$ $= {{10^2} \over {21}} = {k \over {21}}$ Multiply both sides by $21$: $k = 10^2$ $k = 100$

Thus, the value of $k$ is $100$.

3. Tips for Success and Common Pitfalls

• Recognize Binomial Identities: Many problems involving binomial coefficients rely on recognizing and applying identities like Pascal's Identity, Vandermonde's Identity, or properties of symmetry (${}^{n}{C_r} = {}^{n}{C_{n-r}}$). Familiarity with these identities is key.
• Factorial Simplification: Be careful when simplifying expressions involving factorials. Remember that $n! = n \times (n-1)!$ and $(n+1)! = (n+1) \times n!$.
• Summation Formulas: Memorize or be able to quickly derive common summation formulas for powers of natural numbers ($\sum n$, $\sum n^2$, $\sum n^3$). These are frequently tested.
• Algebraic Precision: Even after applying the correct concepts, a small algebraic error can lead to an incorrect answer. Double-check each step of simplification.

4. Summary and Key Takeaway

This problem beautifully demonstrates how a complex-looking summation can be simplified significantly by applying fundamental combinatorial identities and summation formulas. The key steps involved:

1. Using Pascal's Identity to simplify the sum of binomial coefficients in the denominator.
2. Carefully simplifying the ratio of binomial coefficients using their factorial definition.
3. Recognizing the resulting summation as the sum of cubes of natural numbers.
4. Applying the formula for the sum of cubes to evaluate the expression and solve for $k$.

Mastering binomial coefficient identities and common series summation formulas is essential for efficiently solving such problems in JEE Mathematics.