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Question

If limnna=2sin1[(a2+2a)(a1)(a+1)a(a+1)]=kπ120, then the value of k is

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Solution

Sn=na=2sin1[(a2+2a)(a1)(a+1)a(a+1)]

Putting 1a=x,1a+1=y
Let,
L=[(a2+2a)(a1)(a+1)a(a+1)]
=xy[(a+1)21a21]
=xy[1y211x21]
=x1y2y1x2

Ta=sin1(x1y2y1x2)
=sin1xsin1y =sin11asin11a+1
Substituting a=2,3,4,...,n
Sn=sin112sin113 +sin113sin114 +sin114sin115 : : +sin11nsin11n+1

limnSn=limn(sin112sin11n+1)
=π6+0=π6

k=20

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