Key Concept: Sum of Binomial Coefficients The Binomial Theorem states that for any non-negative integer $n$, the expansion of $(x+y)^n$ is given by: $(x+y)^n = \sum_{r=0}^{n} {}^{n}{C_r} x^{n-r} y^r$ A fundamental result derived from this theorem, and crucial for solving this problem, is the sum of all binomial coefficients for a given $n$. By setting $x=1$ and $y=1$, we get: $\sum_{r=0}^{n} {}^{n}{C_r} = {}^{n}{C_0} + {}^{n}{C_1} + {}^{n}{C_2} + \dots + {}^{n}{C_n} = (1+1)^n = 2^n$ Additionally, we use the symmetry property of binomial coefficients: ${}^{n}{C_r} = {}^{n}{C_{n-r}}$ This property implies that the coefficients are symmetric around the middle term(s) of the expansion.

Problem Statement We are asked to evaluate the following expression: $\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right) + \left( {{}^{21}{C_4} - {}^{10}{C_4}} \right) + \dots + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$

Step-by-step Working

1. Re-arranging the Summation The given expression is a sum of differences. We can rewrite this as the difference of two separate summations. This simplification is based on the linearity of summation: $\sum (a_r - b_r) = \sum a_r - \sum b_r$. Let the given expression be $S$. $S = \sum_{r=1}^{10} \left( {}^{21}{C_r} - {}^{10}{C_r} \right)$ $S = \left( \sum_{r=1}^{10} {}^{21}{C_r} \right) - \left( \sum_{r=1}^{10} {}^{10}{C_r} \right)$ Let's evaluate each sum independently.

2. Evaluating the Second Sum: $\sum_{r=1}^{10} {}^{10}{C_r}$ This sum involves binomial coefficients for $n=10$. We know the complete sum for $n=10$ is $2^{10}$. $\sum_{r=0}^{10} {}^{10}{C_r} = {}^{10}{C_0} + {}^{10}{C_1} + {}^{10}{C_2} + \dots + {}^{10}{C_{10}} = 2^{10}$ The sum we need is $S_2 = {}^{10}{C_1} + {}^{10}{C_2} + \dots + {}^{10}{C_{10}}$. This sum includes all terms from $r=1$ up to $r=10$. We can express $S_2$ by subtracting the missing ${}^{10}{C_0}$ term from the complete sum: $S_2 = \left( \sum_{r=0}^{10} {}^{10}{C_r} \right) - {}^{10}{C_0}$ Since ${}^{10}{C_0} = 1$: $S_2 = 2^{10} - 1$

3. Evaluating the First Sum: $\sum_{r=1}^{10} {}^{21}{C_r}$ This sum involves binomial coefficients for $n=21$. The complete sum for $n=21$ is $2^{21}$: $\sum_{r=0}^{21} {}^{21}{C_r} = {}^{21}{C_0} + {}^{21}{C_1} + \dots + {}^{21}{C_{10}} + {}^{21}{C_{11}} + \dots + {}^{21}{C_{21}} = 2^{21}$ We need to find $S_1 = {}^{21}{C_1} + {}^{21}{C_2} + \dots + {}^{21}{C_{10}}$. We use the symmetry property ${}^{n}{C_r} = {}^{n}{C_{n-r}}$. For $n=21$, this means ${}^{21}{C_r} = {}^{21}{C_{21-r}}$. Specifically: ${}^{21}{C_1} = {}^{21}{C_{20}}$ ${}^{21}{C_2} = {}^{21}{C_{19}}$ ... ${}^{21}{C_{10}} = {}^{21}{C_{11}}$ So, the terms from ${}^{21}{C_1}$ to ${}^{21}{C_{10}}$ are identical to the terms from ${}^{21}{C_{11}}$ to ${}^{21}{C_{20}}$ in reverse order. Let $S_1 = {}^{21}{C_1} + \dots + {}^{21}{C_{10}}$. Then, the sum of the terms from $r=11$ to $r=20$ is also $S_1$: $\sum_{r=11}^{20} {}^{21}{C_r} = {}^{21}{C_{11}} + \dots + {}^{21}{C_{20}} = S_1$ Now, substitute these back into the full sum identity: $\sum_{r=0}^{21} {}^{21}{C_r} = {}^{21}{C_0} + \left( \sum_{r=1}^{10} {}^{21}{C_r} \right) + \left( \sum_{r=11}^{20} {}^{21}{C_r} \right) + {}^{21}{C_{21}}$ $2^{21} = {}^{21}{C_0} + S_1 + S_1 + {}^{21}{C_{21}}$ Since ${}^{21}{C_0} = 1$ and ${}^{21}{C_{21}} = 1$: $2^{21} = 1 + 2S_1 + 1$ $2^{21} = 2 + 2S_1$ Subtract 2 from both sides: $2S_1 = 2^{21} - 2$ Divide by 2: $S_1 = \frac{2^{21} - 2}{2} = 2^{20} - 1$

4. Combining the Results Now, substitute the values of $S_1$ and $S_2$ back into the rearranged expression for $S$: $S = S_1 - S_2$ $S = (2^{20} - 1) - (2^{10} - 1)$ $S = 2^{20} - 1 - 2^{10} + 1$ $S = 2^{20} - 2^{10}$

Tips and Common Mistakes

• Starting Index: Always pay close attention to the starting index of the summation. The identity $\sum_{r=0}^{n} {}^{n}{C_r} = 2^n$ starts from $r=0$. If the sum starts from $r=1$, you must subtract ${}^{n}{C_0}$ (which is always 1) from the total sum $2^n$.
• Symmetry for Odd $n$: For an odd $n$ (like $n=21$), the middle term is not a single coefficient but rather the split between two equal halves. The sum of coefficients from $r=0$ to $(n-1)/2$ is equal to $2^{n-1}$. For example, $\sum_{r=0}^{10} {}^{21}{C_r} = 2^{21-1} = 2^{20}$. Therefore, $\sum_{r=1}^{10} {}^{21}{C_r} = \left(\sum_{r=0}^{10} {}^{21}{C_r}\right) - {}^{21}{C_0} = 2^{20} - 1$. This provides an alternative, quicker way to calculate $S_1$.

Summary By breaking down the complex expression into two simpler sums and applying the fundamental properties of binomial coefficients, specifically the sum of all coefficients ($2^n$) and the symmetry property (${}^{n}{C_r} = {}^{n}{C_{n-r}}$), we systematically evaluated each part. The first sum, $\sum_{r=1}^{10} {}^{21}{C_r}$, evaluates to $2^{20} - 1$, and the second sum, $\sum_{r=1}^{10} {}^{10}{C_r}$, evaluates to $2^{10} - 1$. Their difference yields the final result.

The value of the expression is $\mathbf{2^{20} - 2^{10}}$.