ABCD is a square of length a,a∈N,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 a−1∑n=1(AL22+LnM2n) is equal to
A
12a(a−1)2
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B
12(a−1)(2a−1)(4a−1)
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C
12a(a−1)(4a−1)
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D
None of these
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Solution
The correct option is C12a(a−1)(4a−1) (ALn)2=(AB)2+(BLn)2=a2+n2(LnMn)2=(a−n)2+n2a−1∑n=1((ALn)2+(LnMn)2)=a−1∑n=1(a2+n2+n2+n2−2an+a2)a−1∑n=1(3n2−2an+2a2)=3a−1∑n=1n2−2aa−1∑n=1n+2a2a−1∑n=1(1)3(a−1)a(2a−1)6−2a(a−1)a2+2a2(a−1)(a−1)a[(2a−1)2−a+2a]=12a(a−1)(4a−1)