ABCD is a square of length `a,a in N ,a gt 1.` Let `L_1,L_2,L_3,` ……. Be points on BC such that `BL_1L_2=L_2L_3=….=1` and `M_1,M_2M_3….` be points on CD such that `CM_1=M_1M_2 =M_2M_3 =….1` Then `Sigma_(n=1)^(a-1) (AL_n^2+L_nM_n^2)` is equal to

Options

A. `1/2 a(a-1)^2`
B. `1/2(a-1)(2a-1)(4a-1)`
C. `1/2a(a-1)^2`
D. none of these

Answer

C

Explanation

`(AL_(1))^(2)+(L_(1)M_(1))^(2)=(a^(2)+1^(2))+{(a-1)^(2)+1^(2)}`
`(AL_(2))^(2)+(L_(2)M_(2))^(2)=(a^(2)+2^(2))+{(a-2)^(2)+2^(2)}`
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`(AL_(a-1))^(2)+(L_(a-1)M_(a-1))^(2)={a^(2)+(a-1)^(2)}+{1^(2)+(a-1)^(2)}`
Therefore, the required sum is
`(a-1)a^(2)+{1^(2)+2^(2)+..+(a-1)^(2)}+2{1^(2)+2^(2)+..+(a-1)^(2)}`
`=(a-1)a^(2)+3((a-1)a(2a-1))/(6`
`=a(a-1)(a+(2a-1)/2)`
`=1/2a(a-1)(4a-1)`