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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

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The correct Answer is:
1
`A_(k).A_(k+1)=[(a_(k-1), 0),(0,a_(k))][(a_(k),0),(0,a_(k+1))]`
`A_(k).A_(k+1)=[(a_(k-1).a_(k), 0),(0,a_(k)a_(k+1))]`
`A_(k).A_(k+1)=[(.^(n)C_(k-1) .^(n)C_(k), 0),(0, .^(n)C_(k) .^(n)C_(k+1))]`
`B=[(sum_(k=1)^(n=1).^(n)C_(k-1).^(n)C_(k),0),(0,sum_(k=1)^(n-1).^(n)C_(k).^(n)C_(k+1))]=[(a,0),(0,b)]`
`a=b=.^(2n)C_(n-1-n)`
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