<p>If \((a+1)(b+1)(c+1)(d+1) = 1\), \((a+2)(b+2)(c+2)(d+2) = 2\), \((a+3)(b+3)(c+3)(d+3) = 3\), \((a+4)(b+4)(c+4)(d+4) = 4\). Then the value of \((a+5)(b+5)(c+5)(d+5)\) is</p>
Step-by-Step Solution
Key Concept: Define f(x) = (a+x)(b+x)(c+x)(d+x) as a polynomial in x; the given conditions f(1)=1, f(2)=2, f(3)=3, f(4)=4 mean that f(x)-x vanishes at four points, allowing us to determine f(x) completely and find f(5).
<p><strong>Step 1:</strong> Define f(x) = (a+x)(b+x)(c+x)(d+x). This is a monic polynomial of degree 4 (leading term x⁴).</p><p><strong>Step 2:</strong> From given conditions: f(1)=1, f(2)=2, f(3)=3, f(4)=4. This means f(x)−x equals zero at x=1,2,3,4.</p><p><strong>Step 3:</strong> Since f(x)−x is a degree 4 polynomial with four known roots: f(x)−x = k(x−1)(x−2)(x−3)(x−4) for some constant k.</p><p><strong>Step 4:</strong> Expand f(x) = x⁴ + (a+b+c+d)x³ + ... The leading coefficient of f(x)−x is 1 (from x⁴ term). Expand k(x−1)(x−2)(x−3)(x−4) = k[x⁴ − 10x³ + ...]. Comparing leading coefficients: k = 1.</p><p><strong>Step 5:</strong> Therefore: f(x)−x = (x−1)(x−2)(x−3)(x−4). Expanding the right side: (x−1)(x−2)(x−3)(x−4) = x⁴−10x³+35x²−50x+24.</p><p><strong>Step 6:</strong> So f(x) = x⁴−10x³+35x²−50x+24 + x = x⁴−10x³+35x²−49x+24.</p><p><strong>Step 7:</strong> Find f(5) = 625−10(125)+35(25)−49(5)+24 = 625−1250+875−245+24 = 29.</p><p>∴ Answer: D (or 29)</p>
Correct Answer: D