<p>If \((a+1)(b+1)(c+1)(d+1) = 1\)<br>\((a+2)(b+2)(c+2)(d+2) = 2\)<br>\((a+3)(b+3)(c+3)(d+3) = 3\)<br>\((a+4)(b+4)(c+4)(d+4) = 4\)<br>Then find \((a+5)(b+5)(c+5)(d+5)\)</p>
Step-by-Step Solution
Key Concept: Define f(x) = (a+x)(b+x)(c+x)(d+x) as a polynomial of degree 4. Use the given conditions f(1)=1, f(2)=2, f(3)=3, f(4)=4 to recognize that g(x) = f(x) - x has four roots at x=1,2,3,4, allowing us to write g(x) = k(x-1)(x-2)(x-3)(x-4) for some constant k.
<p><strong>Step 1:</strong> Define f(x) = (a+x)(b+x)(c+x)(d+x). This is a degree 4 polynomial with leading coefficient 1.</p><p><strong>Step 2:</strong> Given conditions: f(1)=1, f(2)=2, f(3)=3, f(4)=4. These can be rewritten as:<br>f(1) - 1 = 0, f(2) - 2 = 0, f(3) - 3 = 0, f(4) - 4 = 0</p><p><strong>Step 3:</strong> Consider g(x) = f(x) - x. This is also a degree 4 polynomial with leading coefficient 1. Since g(x) = 0 at x = 1, 2, 3, 4, we can write:<br>g(x) = (x-1)(x-2)(x-3)(x-4) + k for some constant k</p><p><strong>Step 4:</strong> Actually, since f(x) has leading coefficient 1 and g(x) = f(x) - x also has leading coefficient 1, and g(x) vanishes at 4 points:<br>g(x) = (x-1)(x-2)(x-3)(x-4)</p><p><strong>Step 5:</strong> Therefore: f(x) = (x-1)(x-2)(x-3)(x-4) + x</p><p><strong>Step 6:</strong> Find f(5):<br>f(5) = (5-1)(5-2)(5-3)(5-4) + 5<br>f(5) = (4)(3)(2)(1) + 5<br>f(5) = 24 + 5 = 29</p><p><strong>∴ Answer: B (29)</strong></p>
Correct Answer: B