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Question

Let a sequence {an} be defined by an=1n+1+1n+2+1n+3+.....+13n, then

A
a2=712
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B
a2=1920
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C
an+1an=9n+5(3n+1)(3n+2)(3n+3)
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D
an+1an=23(n+1)
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Solution

The correct options are
C a2=1920
D an+1an=9n+5(3n+1)(3n+2)(3n+3)
an is sum of reciprocals of natural numbers starting from n+1 and ending at 3n.
a2=13+14+15+16=1920
an+1=1n+2+1n+3+....+13n+13n+1+13n+2+13n+3
an+1an=13n+1+13n+2+13n+31n+1
=13n+1+13n+223n+3
=(3n+2)(3n+3)(3n+1)(3n+2)(3n+3)+(3n+1)(3n+3)(3n+1)(3n+2)(3n+3)2(3n+1)(3n+2)(3n+1)(3n+2)(3n+3)
=9n+5(3n+1)(3n+2)(3n+3)
Hence, options 'B' and 'C' are correct.

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