Sol.  =

Let y = x –3 ⇒ y + 1 = x –2

So the given expression reduces to =

⇒ 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 for all k ≥ n, then we get

⇒ a 0 + a 1 (1+ y) + a 2 (1+ y)

2 +….+ a n–1 (1+ y) n–1 +(1 + y) n + (1 + y) n+1 +…..+(1+y) 2n =b 0 + b 1 y+ ….+b n y n + …..+b 2n y 2n Compare the coefficients of y

n on both sides, we get

n C n + n+1 C n + n+2 C n +….+ 2n C n = b n

⇒ n+1 C n+1 + n+1 C n + n+2 C n +……+ 2n C n = b n

(use n C r + n C r–1 = n+1 C r )

⇒ n+2 C n+1 + n+2 C n +……+ 2n C n = b n (adding the first two terms)

⇒ If we combine terms on L.H.S. finally we get 2n+1 C n+1 = b n