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Let, C(k) = ""^(n)C(k) " for" 0 le kle n...

Let, `C_(k) = ""^(n)C_(k) " for" 0 le kle n and A_(k) = [[C_(k-1)^(2),0],[0,C_(k)^(2)]]` for `k ge l` and `A_(1) + A_(2) + A_(3) +...+ A_(n) = [[k_(1),0],[0, k_(2)]]`, then

A

`k_(1) = K_(2)`

B

`k_(1) + k_(2) = ""^(2n)C_(2n)+1`

C

`k_(1) = ""^(2n)C_(n)-1`

D

`k_(2) = ""^(2n)C_(n+1)`

Text Solution

Verified by Experts

The correct Answer is:
A , C
`A_(1) + A_(2) + A_(3) +... + A_(n) = [[C_(0)^(2),0],[0,C_(1)^(2)]]+ [[C_(1)^(2),0],[0, C_(2)^(2)]]`
`+ [[C_(2)^(2),0],[0,C_(3)^(2)]]+...+ [[C_(n-1)^(2),0],[0, C_(n)^(2)]]`
` = [[C_(0)^(2)+C_(1)^(2) + C_(2)^(2)+...+C_(n-1)^(2),0],[0,C_(1)^(2)+C_(2)^(2)+C_(2)^(3)+...+C_(n)^(2)]]`
`[[""^(2n)C_(n)-1,0],[0,""^(2n)C_(n)-1]] =[[k_(1),0],[0,k_(2)]]` [given ]
`therefore k_(1) = k_(2) =""^(2n) C_(n)-1`
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