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Q.

sin112+sin1216+sin13212+sin12320+sin15230+....=

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a

π4

b

π12

c

π8

d

π2

answer is B.

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Detailed Solution

=1sin1nn1n(n+1)

=1sin1[1nnn+11n+1n1n]

=1sin1[1n11n+11n+111n]

=Ltn1n[sin11nsin11n+1]

=Ltn[π2sin11n+1]=π2 (by telescopic) 

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