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the value of the determinant`|{:(.^(n)C_(r-1),,.^(n)C_(r),,(r+1)^(n+2)C_(r+1)),(.^(n)C_(r),,.^(n)C_(r+1),,(r+2)^(n+2)C_(r+2)),(.^(n)C_(r+1),,.^(n)C_(r+2),,(r+3)^(n+2)C_(r+3)):}|` isA. `n^(2)+n-1)`B. `0`C. `.^(n+3)C_(r+3)`D. `.^(n)C_(r-1)+^(n)C_(r)+^(n)C_(r+1)` |
Answer» Correct Answer - B `Delta = |{:(.^(n)C_(r-1),,.^(n)C_(r),,(r+1)^(n+2)C_(r+1)),(.^(n)C_(r),,.^(n)C_(r+1),,(n+2)^(n+1)C_(r+1)),(.^(n)C_(r+1),,.^(n)C_(r+2),,(r+3)^(n+2)C_(r+3)):}|` Applying `C_(1) to C_(1)+C_(2) " and using " .^(n)C_(r)=(n)/(r )^(n-1) C_(r+1) " in " C_(3)` we get ` Delta =|{:(.^(n+1)C_(r),,.^(n)C_(r),,(n+2)^(n+1)C_(r)),(.^(n+1)C_(r+1),,.^(n)C_(r+1),,(n+2)^(n+1)C_(r+1)),(.^(n+1)C_(r+2),,.^(n)C_(r+2),,(n+2)^(n+1)C_(r+2)):}|` `=(n+2) |{:(.^(n+1)C_(r),,.^(n)C_(r),,.^(n+1)C_(r)),(.^(n+1)C_(r+1),,.^(n)C_(r+1),,.^(n+1)C_(r+1)),(.^(n+1)C_(r+2),,.^(n)C_(r+2),,.^(n+1)C_(r+2)):}|` ( as `C_(1) " and " C_(3)`are identical ) |
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