If `sum_(r=0)^(2n) a_(r ) (x - 2)^(r ) = sum_(r=0)^(2n) b_(r ) (x - 3)^(r )` and `a_(K) = 1, AA k ge n`, then show that `b_(n) = ""^(2n + 1)C_(n + 1)`

Explanation

Let ` (y= a)^(m)` , where m is a positive integer, r `le` m
Now , ` (dy)/(dx) = m (x - a)^(m-1) rArr (d^(2) y)/(dx^(2)) = m (m-1) (x - a) ^(m -2)`
`rArr (d^(3) y)/(dx^(3)) = m (m -1) (m-2) (m-3) (x -a)^(n-4)`
On differentiating r times , we get
`(d^(r) y)/(dx^(r)) = m (m-1)...(m - r + 1) (x -a)^(m-r)`
`= (m!)/((m-r)!) (x -a)^(m-r) = r !(""^(m)C_(r)) (x - a)^(n-r)`
and for ` r gt m , (d^(r)y)/(dx^(r)) = 0 `
Now , ` sum_(r=0)^(2n) a , (x - 2)^(r) = sum_(r=0)^(2n) b_(r)` [given]
On differnentiating both sides n times w.r.t.x, we get
` sum_(r=0)^(2n) a_(r) (n!)^(r) C_(n) (x - 2)^(r-n) = sum_(r=0)^(2n) b_(r) (n!) ""^(r)C_(n) (x-3) ^(r-n)`
On putting x = 3, we get
[ since ,all the terms except first on RHS bocome zero]
`rArr b_(n) = ""^(n)C_(n) + ""^(n+1)C_(n) + ""^(n+2)C_(n) +...+ ""^(2n)C_(n) `
` [because a_(r) = 1, rArr ge n]`
` (""^(n+2)C_(n+1) + ""^(n+2)C_(n)) +...+ ""^(2n)C_(n)`
`=""^(n+3)C_(n+1)+...+""^(n2)C_(n)=...`
`= ""^(n2)C_(n+1)+ ""^(n2)C_(n)= ""^(2n+1)C_(n+1)`.