Let where the symbols have their usual meanings. The f(n) is divisible by

Engineering

Mathematics

Introduction to Determinants

Question

Let $f (n) = ∣ \begin{array}{ccc} n & n + 1 & n + 2 \\ ^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 2} P_{n + 2} \\ ^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 2} C_{n + 2} \end{array} ∣$ where the symbols have their usual meanings. The f(n) is divisible by

n² + n + 1

(n + 1)!

none of these

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$∣ \begin{array}{ccc} n & n + 1 & n + 2 \\ ^{n} P_{n} & ^{n + 1} P_{n + 1} & ^{n + 2} P_{n + 2} \\ ^{n} C_{n} & ^{n + 1} C_{n + 1} & ^{n + 2} C_{n + 2} \end{array} ∣$
$= ∣ \begin{array}{ccc} n & n + 1 & n + 2 \\ n! & (n + 1)! & (n + 2)! \\ 1 & 1 & 1 \end{array} ∣$

Doing, C₁ = C₁ – C₂ and C₂ = C₂ – C₃,
$= n! ∣ \begin{array}{ccc} - 1 & - 1 & n + 2 \\ - n & - n^{2} - 2 n - 1 & (n + 2) (n + 1) \\ 0 & 0 & 1 \end{array} ∣$

= – 1(– n² – 2n – n – 0) + (– n – 0) + (0)
= n!(n² + 2n + 1 – n)

= n!(n² + n + 1)
So, it is divisible by n! and (n² + n + 1)

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