1. Key Concepts and Formulas

This problem primarily relies on two fundamental identities involving binomial coefficients, which are essential for simplifying the given summation:

• Identity 1: Relationship between $k {^n C_k}$ and ${^{n-1} C_{k-1}}$ The identity states that $k \cdot {^n C_k} = n \cdot {^{n-1} C_{k-1}}$.

  • Why it's useful: This identity is a powerful tool for simplifying expressions where the index $k$ (or a multiple of $k$) appears as a multiplier of a binomial coefficient ${^n C_k}$ within a summation. It allows us to reduce the upper index of the binomial coefficient from $n$ to $n-1$ and effectively removes the multiplying $k$, thereby making the expression more amenable to further algebraic manipulation or the application of other identities.
  • Brief Derivation: $k \cdot {^n C_k} = k \cdot \frac{n!}{k!(n-k)!} = k \cdot \frac{n!}{k \cdot (k-1)!(n-k)!} = \frac{n!}{(k-1)!(n-k)!}$ By factoring out $n$ from $n! = n \cdot (n-1)!$, we get: $k \cdot {^n C_k} = n \cdot \frac{(n-1)!}{(k-1)!(n-k)!} = n \cdot {^{n-1} C_{k-1}}$

• Identity 2: Sum of Squares of Binomial Coefficients The identity states that $\sum\limits_{k=0}^{n} ({^n C_k})^2 = {^{2n} C_n}$.

  • Why it's useful: This identity provides a direct and elegant way to calculate the sum of the squares of all binomial coefficients in a given row ($n$) of Pascal's triangle. It is a special case of Vandermonde's Identity and is frequently encountered in problems involving sums of products of binomial coefficients.
  • Connection to Vandermonde's Identity: Vandermonde's Identity is $\sum\limits_{k=0}^{r} {^m C_k} {^n C_{r-k}} = {^{m+n} C_r}$. By setting $m=n$ and $r=n$, and using the property ${^n C_{n-k}} = {^n C_k}$, the identity transforms to $\sum\limits_{k=0}^{n} {^n C_k} {^n C_{n-k}} = \sum\limits_{k=0}^{n} ({^n C_k})^2 = {^{2n} C_n}$.

2. Step-by-Step Solution

We are tasked with finding the value of $L$ from the given equation: $\sum\limits_{k=1}^{10} k^{2}\left({^{10} C_{k}}\right)^{2} = 22000 L$

Step 1: Simplify the general term $k^2 ({^{10} C_k})^2$ using Identity 1

Let's examine the general term inside the summation: $k^2 ({^{10} C_k})^2$. We can rewrite this term to group $k$ with ${^{10} C_k}$: $k^2 ({^{10} C_k})^2 = (k \cdot {^{10} C_k})^2$ Now, we apply Identity 1 ($k \cdot {^n C_k} = n \cdot {^{n-1} C_{k-1}}$) using $n=10$: $k \cdot {^{10} C_k} = 10 \cdot {^{10-1} C_{k-1}} = 10 \cdot {^9 C_{k-1}}$

• Why this step is taken: The purpose of this substitution is to eliminate the variable $k$ that multiplies the binomial coefficient, which is a common strategy to simplify expressions within summations. By applying Identity 1, we convert the term into a product of a constant (10) and a simpler binomial coefficient (${^9 C_{k-1}}$), making the overall summation easier to evaluate.

Substitute this simplified form back into the squared term: $(k \cdot {^{10} C_k})^2 = (10 \cdot {^9 C_{k-1}})^2 = 10^2 \cdot ({^9 C_{k-1}})^2 = 100 \cdot ({^9 C_{k-1}})^2$

Step 2: Rewrite the summation with the simplified term

Now, we replace the original general term in the summation with our newly derived simplified expression: $\sum\limits_{k=1}^{10} \left[ 100 \cdot ({^9 C_{k-1}})^2 \right] = 22000 L$ Since 100 is a constant factor for every term in the sum, we can factor it out of the summation: $100 \sum\limits_{k=1}^{10} ({^9 C_{k-1}})^2 = 22000 L$

• Why this step is taken: Factoring out constants is a fundamental property of summations that helps to simplify the expression and isolate the core summation that needs to be evaluated.

Step 3: Adjust the index of summation to match Identity 2

To apply Identity 2 ($\sum\limits_{k=0}^{n} ({^n C_k})^2$), which requires the summation to start from $k=0$, we need to adjust the index of our current summation. Let $j = k-1$. When $k=1$, the new index $j = 1-1 = 0$. When $k=10$, the new index $j = 10-1 = 9$. The term ${^9 C_{k-1}}$ becomes ${^9 C_j}$.

The summation now transforms to: $100 \sum\limits_{j=0}^{9} ({^9 C_j})^2 = 22000 L$

• Why this step is taken: This change of variable (index transformation) is critical. It standardizes the form of the summation to directly match the pattern of Identity 2, allowing for its straightforward application. Without this adjustment, Identity 2 cannot be used directly.

Step 4: Apply Identity 2 to evaluate the summation

The summation $\sum\limits_{j=0}^{9} ({^9 C_j})^2$ is now in the exact form of Identity 2, $\sum\limits_{k=0}^{n} ({^n C_k})^2$, where $n=9$. Applying the identity: $\sum\limits_{j=0}^{9} ({^9 C_j})^2 = {^{2 \times 9} C_9} = {^{18} C_9}$

• Why this step is taken: This is the crucial simplification. By recognizing and applying Identity 2, we replace a series of squared binomial coefficients with a single, much simpler binomial coefficient. This dramatically reduces the complexity of the problem from a summation to a single term.

Step 5: Substitute the result and solve for L

Substitute the value of the summation (${^{18} C_9}$) back into our equation: $100 \cdot {^{18} C_9} = 22000 L$ Now, we isolate $L$: $L = \frac{100 \cdot {^{18} C_9}}{22000}$ $L = \frac{{^{18} C_9}}{220}$

To find the numerical value of $L$, we first calculate ${^{18} C_9}$: ${^{18} C_9} = \frac{18!}{9! \cdot (18-9)!} = \frac{18!}{9! \cdot 9!}$ ${^{18} C_9} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$ Performing the calculation: ${^{18} C_9} = 48620$

• Why this step is taken: This is the final arithmetic evaluation. After simplifying the combinatorial expression, we compute its numerical value and substitute it back to find the unknown $L$.

Finally, substitute the value of ${^{18} C_9}$ into the expression for $L$: $L = \frac{48620}{220}$ $L = 221$

3. Tips for Success and Common Mistakes

• Recognize Binomial Identities: The most important skill here is to identify when and how to apply standard binomial identities. Memorizing and understanding the derivations of key identities will save significant time.
• Careful Index Manipulation: When changing the index of summation (e.g., from $k$ to $j=k-1$), ensure that you correctly adjust both the variable within the sum and the upper and lower limits of the summation. A common error is failing to change the lower limit from $k=1$ to $j=0$, which would omit the first term and lead to an incorrect result.
• Algebraic Precision: Meticulously perform algebraic manipulations, such as factoring out constants and handling squares. Small errors in these steps can propagate and lead to incorrect answers.
• Factorial Calculations: Be systematic when calculating factorial expressions like ${^n C_k}$ to avoid arithmetic mistakes.
• Practice with Variations: Many problems will combine these identities in slightly different ways. Practice with variations to build flexibility in applying these concepts.

4. Summary and Key Takeaway

This problem is an excellent illustration of how two fundamental binomial identities can be sequentially applied to simplify a complex summation problem into a straightforward calculation. The process involved:

1. Using $k \cdot {^n C_k} = n \cdot {^{n-1} C_{k-1}}$ to simplify the term $k \cdot {^{10} C_k}$.
2. Adjusting the summation index to align with the standard form of the sum of squares identity.
3. Applying $\sum\limits_{k=0}^{n} ({^n C_k})^2 = {^{2n} C_n}$ to evaluate the sum.
4. Performing the final arithmetic to solve for $L$.

The key takeaway is that by strategically transforming expressions and recognizing underlying combinatorial patterns, seemingly difficult problems can be reduced to manageable steps.