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Question

ABCD is a square of length a,aN,a>1. Let L1,L2,L3......, be point on BC such that BL1=L1L2=L2L3=.....=1 and M1,M2,M3......... be points on CD such that CM1=M1M2=...........=1. Then a1n=1(AL22+LnM2n) is equal to

A
12a(a1)2
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B
12(a1)(2a1)(4a1)
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C
12a(a1)(4a1)
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D
None of these
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Solution

The correct option is C 12a(a1)(4a1)
(ALn)2=(AB)2+(BLn)2=a2+n2(LnMn)2=(an)2+n2a1n=1((ALn)2+(LnMn)2)=a1n=1(a2+n2+n2+n22an+a2)a1n=1(3n22an+2a2)=3a1n=1n22aa1n=1n+2a2a1n=1(1)3(a1)a(2a1)62a(a1)a2+2a2(a1)(a1)a[(2a1)2a+2a]=12a(a1)(4a1)

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