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Question

# For a fixed positive integer n, if D=âˆ£âˆ£ âˆ£ âˆ£âˆ£n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!âˆ£âˆ£ âˆ£ âˆ£âˆ£, then [D(n!)3âˆ’4] is___

A
divisible by n.
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B
divisible by n+1
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C
divisible by n+2
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D
divisible by n+3
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Solution

## The correct option is A divisible by n.Given, D=∣∣ ∣ ∣∣n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!∣∣ ∣ ∣∣ Taking n!, (n+1)!and (n+2)! common from R1,R2 and R3, respectively. ∴ D=n!(n+1)!(n+2)!∣∣ ∣ ∣∣1(n+1)(n+1)(n+2)1(n+2)(n+2)(n+3)1(n+3)(n+3)(n+4)∣∣ ∣ ∣∣ Applying R2→R2−R1 and R3→R3−R2, we get D=n!(n+1)!(n+2)!∣∣ ∣∣1(n+1)(n+1)(n+2)012n+4012n+6∣∣ ∣∣ Expanding along C1, we get D = (n!) (n+ 1)! (n + 2)! [(2n + 6) - (2n + 4)] D = (n!) (n + 1)! (n + 2)! [2] On dividing both side by (n!)3 ⇒ D(n!)3=(n!)(n!)(n+1)(n!)(n+1)(n+2)2(n!)3⇒ D(n!)3=2(n+1)(n+1)(n+2)⇒ D(n!)3=2(n3+4n2+5n+2)=2n(n2+4n+5)+4⇒ D(n!)3−4=2n(n2+4n+5) which shows that [D(n!)3−4] is divisible by n.

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