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the value of the determinant |{:((a(1...

the value of the determinant
`|{:((a_(1)-b_(1))^(2),,(a_(1)-b_(2))^(2),,(a_(1)-b_(3))^(2),,(a_(1)-b_(4))^(2)),((a_(2)-b_(1))^(2),,(a_(2)-b_(2))^(2) ,,(a_(2)-b_(3))^(2),,(a_(3)-b_(4))^(2)),((a_(3)-b_(1))^(2),,(a_(3)-b_(2))^(2),,(a_(3)-b_(3))^(2),,(a_(3)-b_(4))^(2)),((a_(4)-b_(1))^(2),,(a_(4)-b_(2))^(2),,(a_(4)-b_(3))^(2),,(a_(4)-b_(4))^(2)):}|` is

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Answer each question by selecting the proper alternative from those given below each question so as to make the statement true: The solution of the pair of linear equations a_(1)x+b_(1)y+c_(1)=0 and a_(2)x+b_(2)y+c_(2)=0 by corss-multiplication method is given by .................. x=(b_(2)c_(1)-b_(1)c_(2))/(a_(1)b_(2)-a_(2)b_(1)), y=(a_(1)c_(2)-a_(2)c_(1))/(a_(1)b_(2)-a_(2)b_(1)) x=(b_(1)c_(2)-b_(2)c_(1))/(a_(1)b_(2)-a_(2)b_(1)), y=(a_(1)c_(2)-a_(2)c_(1))/(a_(1)b_(2)-a_(2)b_(1)) x=(b_(2)c_(1)-b_(1)c_(2))/(a_(1)b_(2)-a_(2)b_(1)), y=(a_(2)c_(1)-a_(1)c_(2))/(a_(1)b_(2)-a_(2)b_(1)) x=(b_(1)c_(2)-b_(2)c_(1))/(a_(1)b_(2)-a_(2)b_(1)), y=(a_(2)c_(1)-a_(1)c_(2))/(a_(1)b_(2)-a_(2)b_(1))

Statement -1 Consider the determinant Delta=|{:(a_(1)+b_(1)x^(2),a_(1)x^(2)+b_(1),c_(1)),(a_(2)+b_(2)x^(2),a_(2)x^(2)+b_(2),c_(2)),(a_(3)+b_(3)x^(2),a_(3)x^(2)+b_(3),c_(3)):}|=0, where a_(i),b_(i),c_(i) in R (i=1,2,3) and x in R Stement -2 If |{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}| =0, then Delta =0

(1+x)^(10)=a_(0)+a_(1)x+a_(2)x^(2)+……..+a_(10)x^(10) then the value of (a_(0)-a_(2)+a_(4)-a_(6)+a_(8)-a_(10))^(2)+(a_(1)-a_(3)+a_(5)-a_(7)+a_(9))^(2) is ……

If a_(m) be the mth term of an AP, show that a_(1)^(2)-a_(2)^(2)+a_(3)^(2)-a_(4)^(2)+"...."+a_(2n-1)^(2)-a_(2n)^(2)=(n)/((2n-1))(a_(1)^(2)-a_(2n)^(2)) .

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If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2)))=(a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .

If a_(1),b_(1),c_(1),a_(2),b_(2),c_(2) "and" a_(3),b_(3),c_(3) are three digit even natural numbers and Delta=|{:(c_(1),a_(1),b_(1)),(c_(2),a_(2),b_(2)),(c_(3),a_(3),b_(3)):}| , then Delta is