If `sum_(r=0)^(2n) a_(r)(x-1)^(r) = sum_(r=0)^(2n) b_(r)(x-2)^(r)` and `b_(r) = (-1)^(r-n)` for all `r ge n`, then `a_(n)`=A. `.^(2n+1)C_(n-1)`B. `.^(3n)C_(n)`C. `.^(2n+1)C_(n)`D. 0

If `sum_(r=0)^(2n) a_(r)(x-1)^(r) = sum_(r=0)^(2n) b_(r)(x-2)^(r)` and

1.	If `sum_(r=0)^(2n) a_(r)(x-1)^(r) = sum_(r=0)^(2n) b_(r)(x-2)^(r)` and `b_(r) = (-1)^(r-n)` for all `r ge n`, then `a_(n)`=A. `.^(2n+1)C_(n-1)`B. `.^(3n)C_(n)`C. `.^(2n+1)C_(n)`D. 0
Answer» Correct Answer - A

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If `sum_(r=0)^(2n) a_(r)(x-1)^(r) = sum_(r=0)^(2n) b_(r)(x-2)^(r)` and `b_(r) = (-1)^(r-n)` for all `r ge n`, then `a_(n)`=A. `.^(2n+1)C_(n-1)`B. `.^(3n)C_(n)`C. `.^(2n+1)C_(n)`D. 0

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