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

If `sum_(r=0)^(2n) a_(r) (x-2)^(r) = sum_(r=0)^(2n) b_(r),(x-3)^(r)` a

1.	If `sum_(r=0)^(2n) a_(r) (x-2)^(r) = sum_(r=0)^(2n) b_(r),(x-3)^(r)` and `a_(k) = 1` for all `k ge n`, then `b_(n)` is equal toA. `.^(2n+1)C_(n-1)`B. `.^(2n)C_(n+1)`C. `.^(2n)C_(n)`D. `.^(2n+1)C_(n+1)`
Answer» Correct Answer - D In the given equation, put `x - 3 = y`. `:. underset(r=0)overset(2n)suma_(r)(1+y)^(r) = underset(r=0)overset(2n)sumb_(r)(y)^(r)` `rArr a_(0) + a_(1)(1+y) + "….."+ a_(n+1)(1+y)^(n-1)+(1+y)^(n) + (1+y)^(n+1)+"…."+(1+y)^(2n)` `= underset(r=0)overset(2n)sumb_(r)y^(r)` [Using `a_(k) = 1, AA k ge n`] Equating the coeficients of `y^(n)` on both sides, we get `.^(n)C_(n) + .^(n+1)C_(n) + .^(n+2)C_(n) + "......" + .^(2n)C_(n) = b_(n)` `rArr (.^(n+1)C_(n+1)+.^(n+1)C_(n)) + .^(n+2)C_(n) + "...." + .^(2n)C_(n) = b_(n)` [Using `.^(n)C_(n) = .^(n+1)C_(n+1) = 1`] `rArr b_(n) = .^(n+2)C_(n+1) + .^(n+2)C_(n) + "...." + .^(2n)C_(n)` Combing the term in similar way, we get `rArr b_(n) =.^(2n)C_(n+1) + .^(2n)C_(n) = .^(2n+1)C_(n+1)`.

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

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