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 `kgeqn ,` then show that `b_n=^(2n+1)C_(n+1)` .

Explanation

` because sum_(r=0)^(2n) a_(r) (x -2)^(2) = sum_(r=0)^(2n) b_(r) (x-3)^(r)`
Let ` y = x - 3 rArr y + 1 = x - 2`
So, the given expanssion reduces to
`sum_(r=0)^(2n) a_(r) (1 + y)^(r) = sum_(r=0)^(2n) b_(r) y^(r)`
`rArr a_(0) + a_(1) (1 + y) + a_(2) (1 + y)^(2) + ...+ a_(2n) (1 + y)^(2n) `
` = b_(0) + b_(1)y + ...+ b_(2n) y^(2n)`
Using ` a_(k) = 1 ,AA ge n ` , we get
` a_(0) + a_(1) (1 + y) + a_(2) (1 + y)^(2) + ...+ a_(n-1) (1 + y)^(n-2)`
` + (1 + y)^(n) + (1 + y)^(n+1) + ...+ (1 + y)^(2n)`
` = b_(n) + b_(1)y + ...+ b_(n) y^(n) + ...+ b_(2n) y^(2n)`
On comparing the coefficient 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)[because ""^(n)C_(r) + ""^(n)C_(r-1)= ""^(n+1)C_(r)]`
`rArr ""^(n+2)C_(n+1)+ ""^(n+2)C_(n) + ...+ ""^(2n)C_(n) = b_(n)` [adding first two terms]
If we combine terms on LHS finally, we get
` ""^(2n+1)C_(n-1) = b_(n)`