Show that the middle term in the expansion of x - 1 x 2 n is 1 × 3 × 5 × … ( 2 n - 1 ) n ! × ( - 2 ) n .

Given expression is x - 1 x 2 n Since the power of expression is even, we have only one middle term ∴ Middle term = 2 n 2 + 1 t h term i.e., ( n + 1 ) th term. We know that the r + 1 th term in an expansion a + b n is given by T r + 1 = C r n ( a ) n - r ( b ) r ∴ n + 1 th term in the expansion of given expression is T n + 1 = C n 2 n ( x ) 2 n - n - 1 x n = C n 2 n ( x ) n ( - 1 ) n · 1 x n = ( - 1 ) n · C n 2 n = ( - 1 ) n · 2 n ! n ! ( 2 n - n ) ! = ( - 1 ) n · 2 n ! n ! · n ! = ( - 1 ) n · 2 n ( 2 n - 1 ) ( 2 n - 2 ) ( 2 n - 3 ) · … · 1 n ! · n ( n - 1 ) ( n - 2 ) ( n - 3 ) · … · 2 · 1 = ( - 1 ) n 2 n · ( 2 n - 1 ) · 2 ( n - 1 ) ( 2 n - 3 ) · … · 1 n ! · n ( n - 1 ) ( n - 2 ) ( n - 3 ) · … · 1 = ( - 1 ) n · 2 n · [ n ( n - 1 ) ( n - 2 ) ⋯ ] · [ ( 2 n - 1 ) · ( 2 n - 3 ) ⋯ 5 · 3 · 1 ] n ! n ( n - 1 ) ( n - 2 ) ⋯ 1 = ( - 2 ) n [ ( 2 n - 1 ) ( 2 n - 3 ) ⋯ 5 · 3 · 1 ] n ! = 1 × 3 × 5 × ⋯ ( 2 n - 1 ) n ! × ( - 2 ) n Hence, the middle term = 1 × 3 × 5 × ⋯ ( 2 n - 1 ) n ! × ( - 2 ) n

If x p occurs in the expansion of x 2 + 1 x 2 n , then prove that its coefficient is 2 n ! 4 n - p 3 ! 2 n + p 3 !

Given expression is x 2 + 1 x 2 n We know that the general term in the expansion of a + b n is given by T r + 1 = C r n a n - r b r ∴ The general term in the expansion of given expression is T r + 1 = C r 2 n x 2 2 n - r · 1 x r = C r 2 n ( x ) 4 n - 2 r · 1 x r = C r 2 n ( x ) 4 n - 2 r - r = C r 2 n ( x ) 4 n - 3 r If x p occurs in x 2 + 1 x 2 n Then 4 n - 3 r = p ⇒ r = 4 n - p 3 ∴ Coefficient of x p = C r 2 n = C 4 n - p 3 2 n = ( 2 n ) ! 4 n - p 3 ! 2 n - 4 n - p 3 ! = ( 2 n ) ! 4 n - p 3 ! 6 n - 4 n + p 3 ! = ( 2 n ) ! 4 n - p 3 ! 2 n + p 3 ! Hence, the coefficient of x p = ( 2 n ) ! 4 n - p 3 ! 2 n + p 3 ! .