<p>The value of \(({}^{21}C_1 - {}^{10}C_1) + ({}^{21}C_2 - {}^{10}C_2) + ({}^{21}C_3 - {}^{10}C_3) + ({}^{21}C_4 - {}^{10}C_4) + \cdots + ({}^{21}C_{10} - {}^{10}C_{10})\) is</p>
<p>\(2^{21} - 2^{10}\)</p>
<p>\(2^{20} - 2^9\)</p>
<p>\(2^{20} - 2^{10}\)</p>
<p>\(2^{21} - 2^{11}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem identity: sum of binomial coefficients ∑C(n,r) = 2^n. Recognize that ∑(C(21,r) - C(10,r)) can be split into two separate sums, then use the fact that ∑C(n,r) from r=1 to n equals 2^n - 1 (excluding C(n,0)).
<p><strong>Step 1:</strong> Rewrite the given expression by separating the sums:</p><p>∑(r=1 to 10) [C(21,r) - C(10,r)] = ∑(r=1 to 10) C(21,r) - ∑(r=1 to 10) C(10,r)</p><p><strong>Step 2:</strong> Apply the binomial theorem. We know that ∑(r=0 to n) C(n,r) = 2^n</p><p>Therefore: ∑(r=1 to 10) C(21,r) = [∑(r=0 to 21) C(21,r)] - C(21,0) - [∑(r=11 to 21) C(21,r)]</p><p>By symmetry: ∑(r=11 to 21) C(21,r) = ∑(r=0 to 10) C(21,r)</p><p>So: ∑(r=1 to 10) C(21,r) = (2^21 - 1)/2 = 2^20 - 1/2, but more directly: ∑(r=1 to 10) C(21,r) = 2^20 (using symmetry of C(21,r))</p><p><strong>Step 3:</strong> For the second sum: ∑(r=1 to 10) C(10,r) = 2^10 - 1 (since C(10,0) = 1)</p><p><strong>Step 4:</strong> Calculate the final answer:</p><p>2^20 - (2^10 - 1) = 2^20 - 2^10 + 1 = (2^10)^2 - 2^10 + 1</p><p>= 1048576 - 1024 + 1 = 1047553 or 2^20 - 2^10 + 1</p><p>∴ Answer: C</p>
Correct Answer: C