If 1 + x n = C 0 + C 1 x + C 2 x 2 + … … + C n x n , prove that C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = 0 or - 1 n 2 C n 2

1 + x n = ∑ r = 0 n C r x r & 1 - x n = ∑ r = 0 n C r - x r So, 1 + x n 1 - x n = ∑ r = 0 n C r x r × ∑ r = 0 n C r - x r ⇒ 1 - x 2 n = ∑ r = 0 n C r - 1 r x 2 r Compare the coefficient of x n both sides, C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = 0 , if n is odd C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = - 1 n 2 C n 2 , if n is even.

If 1 + x n = C 0 + C 1 x + C 2 x 2 + … … + C n x n , prove that C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = 0 or - 1 n 2 C n 2

1 + x n = ∑ r = 0 n C r x r & 1 - x n = ∑ r = 0 n C r - x r So, 1 + x n 1 - x n = ∑ r = 0 n C r x r × ∑ r = 0 n C r - x r ⇒ 1 - x 2 n = ∑ r = 0 n C r - 1 r x 2 r Compare the coefficient of x n both sides, C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = 0 , if n is odd C 0 2 - C 1 2 + C 2 2 - C 3 2 + … … + - 1 n C n 2 = - 1 n 2 C n 2 , if n is even.

Find the coefficient of x 12 in 4 + 2 x - x 2 1 + x 3 .

4 + 2 x - x 2 1 + x 3 = 4 1 + x - 3 + 2 x 1 + x - 3 - x 2 1 + x - 3 Taking coefficient of x 10 from the first expansion, coefficient of x 9 from the second expansion and coefficient of x 8 from the third expansion we get − 3 − 3 − 1 − 3 − 2 − 3 − 3 − 3 − 4 − 3 − 5 − 3 − 6 − 3 − 7 − 3 − 8 − 3 − 9 10 ! + 2 − 3 − 3 − 1 − 3 − 2 − 3 − 3 − 3 − 4 − 3 − 5 − 3 − 6 − 3 − 7 − 3 − 8 9 ! − − 3 − 3 − 1 − 3 − 2 − 3 − 3 − 3 − 4 − 3 − 5 − 3 − 6 − 3 − 7 8 ! = 4 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 10 ! + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 − 11 9 ! − + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 8 ! = 4 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 - 2 × 11 × 12 − 10 × 11 − 9 × 5 = 264 − 110 − 45 = 109

Find the coefficient of x 100 in 3 - 5 x 1 - x 2 .

3 − 5 x 1 − x - 2 = 1 + 2 x + − 2 − 2 − 1 x 2 2 ! … . . + − 2 − 2 − 1 . . − 2 − 98 x 99 99 ! + − 2 − 2 − 1 … − 2 − 99 x 100 100 ! Coefficient of x 100 is 3 × 101 ! 100 ! - 5 × 100 ! 99 ! = - 197 .

Assuming x to be so small that x 2 and higher powers of x can be neglected. Show that 1 + 3 4 x - 4 16 - 3 x 1 2 8 + x 2 3 is approximately equal to 1 - 305 96 x .

As x is so small that x 2 and higher powers of x can be neglected, we can approximate 1 + x n ≈ 1 + n x Here, we have to approximate the value of 1 + 3 4 x - 4 16 - 3 x 1 2 ( 8 + x ) 2 3 = 1 + 3 4 x - 4 16 1 2 1 - 3 16 x 1 2 ( 8 ) 2 3 1 + x 8 2 3 = 1 + 3 x 4 - 4 1 - 3 x 16 1 2 1 + x 8 - 2 3 ≈ 1 + - 4 3 4 x 1 + 1 2 - 3 x 16 1 + - 2 3 x 8 ≈ 1 - 3 x 1 - 3 32 x 1 - x 12 ≈ 1 - 3 x - 3 32 x 1 - x 12 on neglecting x 2 term ≈ 1 - 99 32 x 1 - x 12 ≈ 1 - x 12 - 99 32 x On neglecting x 2 and higher powers of x ≈ 1 - 305 96 x .