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Question

For all natural numbers n, (n+1)(n+2)(n+3) is divisible by


A

6

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B

5

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C

7

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D

4

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Solution

The correct option is A

6


P(n):(n+1)(n+2)(n+3)P(1):2×3×4=24divisible by 6P(n):(n+1)(n+2)(n+3) is divisible by 6

P(1) is true. Assume P(k) is true
(k+1)(k+2)(k+3)=6mP(k+1):(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3(k+2)(k+3)

Since (k+2) and (k+3) are two consecutive natural numbers, one of them must be even i.e. their product is even, say 2q
(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3.2q=6m+6q=6(m+q)divisible by 6P(k+1) is trueP(n) is true


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