If `sum_(r=0)^(2n)a_r(x-2)^r=sum_(r=0)^(2n)b_r(x-3)^ra n da_k=1` for all `kgeqn ,` then show that `b_n=^(2n+1)C_(n+1)` .

Options

A. `.^(2n+1)C_(n-1)`
B. `.^(2n)C_(n+1)`
C. `.^(2n)C_(n)`
D. `.^(2n+1)C_(n+1)`

Answer

D

Explanation

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)`.