Find the 10 t h term in the expansion of 2 x 2 + 1 x 12 .

Given, 2 x 2 + 1 x 12 ( r + 1 ) t h term in the expansion of ( a + b ) n is C r n a n - r b r . ∴ 10 t h term in the expansion of 2 x 2 + 1 x 12 = C 9 12 2 x 2 12 - 9 1 x 9 = 12 ! 913 ! · 2 3 x 6 · x - 9 ∵ C r n = n ! r ! ( n - r ) ! = 12 × 11 × 10 3 × 2 × 1 · 8 x - 3 ∴ 4 t h term in the expansion of 2 x 2 + 1 x 12 = 1760 x - 3 .

Find 6 th term in the expansion of 4 x 5 - 5 2 x 9

General term of the binomial expansion of a + b n is T r + 1 = C r n a n - r b r where 0 ≤ r < n . Hence, general term of 4 x 5 - 5 2 x 9 is T r + 1 = C r 9 4 x 5 9 - r - 5 2 x r . Now, T 6 = T 5 + 1 = C 5 9 4 x 5 9 - 5 - 5 2 x 5 ⇒ T 6 = - C 5 9 4 x 5 4 5 2 x 5 ⇒ T 6 = - C 5 9 2 8 2 5 5 5 5 4 1 x ⇒ T 6 = - 9 ! 5 ! · 4 ! 40 x ⇒ T 6 = - 5040 x

Find the middle terms in the expansion of 4 x - x 3 8 7 .

Given, 4 x - x 3 8 7 We know that middle terms in the expansion of a + b n are n + 1 2 th and n + 3 2 th term, if n is odd. Therefore, middle term in the expansion of 4 x - x 3 8 7 is 7 + 1 2 th = 4 th and 7 + 3 2 th = 5 th term We know that the r + 1 th term in the expansion of a + b n is T r + 1 = C r n a n - r b r ∴ Middle terms are T 4 = T 3 + 1 = C 3 7 4 x 7 - 3 - x 3 8 3 ⇒ T 4 = C 3 7 2 8 x 4 - x 9 2 9 ⇒ T 4 = - 7 ! 3 ! · 4 ! 1 2 x 13 ⇒ T 4 = - 35 2 x 13 And, T 5 = T 4 + 1 = C 4 7 4 x 7 - 4 - x 3 8 4 ⇒ T 5 = C 4 7 4 3 x 3 x 12 2 12 ⇒ T 5 = 7 ! 3 ! · 4 ! 2 6 x 15 2 12 ⇒ T 5 = 35 64 x 15

Find the coefficient of 1 y 2 in the expansion of y + c 3 y 2 10 .

General term of the binomial expansion of a + b m is T r + 1 = C r m a m - r b r where 0 ≤ r ≤ m . Hence, general term of y + c 3 y 2 10 is T r + 1 = C r 10 y 10 - r c 3 y 2 r ⇒ T r + 1 = C r 10 y 10 - r c 3 r y 2 r ⇒ T r + 1 = C r 10 c 3 r 1 y 3 r - 10 For coefficient of 1 y 2 , put 3 r - 10 = 2 ⇒ 3 r = 12 ⇒ r = 4 . Hence, coefficient of 1 y 2 is = C 4 10 c 12 = 10 ! 6 ! · 4 ! c 12 = 210 c 12