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Question

For a fixed positive integer n, if
D=∣ ∣ ∣n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!∣ ∣ ∣, then [D(n!)34] 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 R2R2R1 and R3R3R2, 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!)34=2n(n2+4n+5)
which shows that [D(n!)34] is divisible by n.

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