Elaborate Solution to Binomial Summation Problem

1. Key Concept: Pascal's Identity (Hockey-stick Identity)

The core property used in solving this problem is Pascal's Identity, which states: ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$ This identity allows us to combine two adjacent binomial coefficients into a single one. Repeated application of this identity forms the basis of the Hockey-stick Identity (also known as the Christmas Stocking Identity): $\sum_{i=r}^{n} {}^{i}{C_r} = {}^{n+1}{C_{r+1}}$ We will use the iterative application of Pascal's Identity to simplify the given sum.

2. Problem Statement

We need to evaluate the sum: $S = \left( {{}^{40}{C_0}} \right) + \left( {{}^{41}{C_1}} \right) + \left( {{}^{42}{C_2}} \right) + \,\,.....\,\, + \,\,\left( {{}^{60}{C_{20}}} \right)$ and express it in the form ${m \over n}{}^{60}{C_{20}}$ where $m$ and $n$ are coprime integers. Finally, we need to find the value of $m+n$.

3. Step-by-Step Solution

Let's rewrite the given sum: $S = {}^{40}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}}$

Step 3.1: Initial Transformation to Enable Pascal's Identity The first term in our sum is ${}^{40}{C_0}$. To begin applying Pascal's Identity, we need two terms with the same upper index ($n$). We know that ${}^n{C_0} = 1$ for any non-negative integer $n$. Therefore, ${}^{40}{C_0} = 1$ and ${}^{41}{C_0} = 1$. We can replace ${}^{40}{C_0}$ with ${}^{41}{C_0}$ without changing the value of the sum. This is a critical step to initiate the telescoping sum. $S = \mathbf{\left( {{}^{41}{C_0}} \right)} + {}^{41}{C_1} + {}^{42}{C_2} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}}$

Step 3.2: Iterative Application of Pascal's Identity

Now we can apply the identity ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$ repeatedly.

• First Combination: Combine the first two terms: ${}^{41}{C_0} + {}^{41}{C_1}$ Using Pascal's Identity with $n=41, r=0$: ${}^{41}{C_0} + {}^{41}{C_1} = {}^{41+1}{C_{0+1}} = {}^{42}{C_1}$ The sum now effectively becomes: $S = \mathbf{\left( {{}^{42}{C_1}} \right)} + {}^{42}{C_2} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}}$

• Second Combination: Combine the new first term with the next term in the original sequence: ${}^{42}{C_1} + {}^{42}{C_2}$ Using Pascal's Identity with $n=42, r=1$: ${}^{42}{C_1} + {}^{42}{C_2} = {}^{42+1}{C_{1+1}} = {}^{43}{C_2}$ The sum now becomes: $S = \mathbf{\left( {{}^{43}{C_2}} \right)} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}}$

• Continuing the Pattern: Notice a clear pattern here: each time we apply the identity, both the upper index and the lower index of the resulting binomial coefficient increment by 1. The result of combining terms ${}^{k}{C_{j}} + {}^{k}{C_{j+1}}$ is ${}^{k+1}{C_{j+1}}$. This process continues all the way until the last term ${}^{60}{C_{20}}$ is combined. The sequence of combinations will be: ${}^{41}{C_0} + {}^{41}{C_1} \longrightarrow {}^{42}{C_1}$ ${}^{42}{C_1} + {}^{42}{C_2} \longrightarrow {}^{43}{C_2}$ ${}^{43}{C_2} + {}^{43}{C_3} \longrightarrow {}^{44}{C_3}$ ... This continues until we have combined ${}^{59}{C_{19}}$ (the result of the previous step) with ${}^{60}{C_{20}}$. The term just before ${}^{60}{C_{20}}$ in the original sum was ${}^{59}{C_{19}}$. So the last pair to combine will be ${}^{60}{C_{19}} + {}^{60}{C_{20}}$ (where ${}^{60}{C_{19}}$ is the cumulative sum of all terms before ${}^{60}{C_{20}}$). Applying Pascal's Identity one final time: ${}^{60}{C_{19}} + {}^{60}{C_{20}} = {}^{60+1}{C_{19+1}} = {}^{61}{C_{20}}$ Therefore, the entire sum simplifies to: $S = {}^{61}{C_{20}}$

Step 3.3: Equating and Solving for m/n

We are given that $S = {m \over n}{}^{60}{C_{20}}$. So, we can write: ${}^{61}{C_{20}} = {m \over n}{}^{60}{C_{20}}$

Now, we expand the binomial coefficients using the definition ${}^N{C_K} = {{N!} \over {K!(N-K)!}}$: ${{61!} \over {20!(61-20)!}} = {m \over n} {{60!} \over {20!(60-20)!}}$ ${{61!} \over {20!41!}} = {m \over n} {{60!} \over {20!40!}}$

We can cancel the common term $20!$ from both sides: ${{61!} \over {41!}} = {m \over n} {{60!} \over {40!}}$

Next, we use the property of factorials: $k! = k \times (k-1)!$ So, $61! = 61 \times 60!$ And $41! = 41 \times 40!$

Substitute these into the equation: ${{61 \times 60!} \over {41 \times 40!}} = {m \over n} {{60!} \over {40!}}$

Now, we can cancel $60!$ and $40!$ from both sides: ${{61} \over {41}} = {m \over n}$

Step 3.4: Finding m+n

We have ${m \over n} = {{61} \over {41}}$. Given that $m$ and $n$ are coprime, and 61 and 41 are both prime numbers (and thus coprime), we can directly assign: $m = 61$ $n = 41$

Finally, we calculate $m+n$: $m+n = 61 + 41 = 102$

4. Tips and Common Mistakes

• Initial Transformation: The most crucial step is often the first one: realizing that ${}^{40}{C_0}$ needs to be cleverly replaced by ${}^{41}{C_0}$ to kickstart the chain of Pascal's Identity applications. Without this, the identity cannot be applied.
• Hockey-stick Identity Recognition: For those familiar, this problem can be solved more quickly by first rewriting the terms using ${}^n{C_r} = {}^n{C_{n-r}}$ to match the form of the Hockey-stick Identity. The sum $S = {}^{40}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \dots + {}^{60}{C_{20}}$ can be written as $S = {}^{40}{C_{40}} + {}^{41}{C_{40}} + {}^{42}{C_{40}} + \dots + {}^{60}{C_{40}}$. Applying $\sum_{i=r}^{n} {}^{i}{C_r} = {}^{n+1}{C_{r+1}}$ directly with $r=40$ and upper limit $n=60$: $S = {}^{60+1}{C_{40+1}} = {}^{61}{C_{41}}$. Then, ${}^{61}{C_{41}} = {{61!} \over {41!20!}} = {{61 \times 60!} \over {41 \times 40!20!}} = {{61} \over {41}} \times {{60!} \over {40!20!}} = {{61} \over {41}}{}^{60}{C_{20}}$. This immediately gives $m=61, n=41$. Both methods yield the same result.
• Factorial Simplification: Be careful when expanding and cancelling factorial terms. Remember that $N! = N \times (N-1)!$.

5. Summary and Key Takeaway

This problem is an excellent illustration of how binomial identities, particularly Pascal's Identity, can be used to simplify complex sums of binomial coefficients. The key lies in recognizing patterns and sometimes making an initial strategic transformation (like changing ${}^{40}{C_0}$ to ${}^{41}{C_0}$) to set up the problem for an elegant solution. Understanding both the iterative application of Pascal's Identity and the direct Hockey-stick Identity provides flexibility in problem-solving. The final steps involve careful algebraic manipulation of factorial expressions to extract the desired values.