( a + b x ) ( 1 - x ) 6 = 3 - 20 x + c x 2 + … then find a , b , c

Given that ( a + b x ) ( 1 - x ) 6 Applying binomial theorem for ( a + b ) n , we get ( a + b ) n = n C 0 · a n + C 1 n · a n - 1 · b + … … + C n - 1 n a · b n - 1 + C n n · b n where C r n = n ! r ! ( n - r ) ! = n · ( n - 1 ) · ( n - 2 ) · … · · r te r m s r ! ( a + b x ) ( 1 - x ) 6 = ( a + b x ) C 0 6 ( 1 ) 6 - C 1 6 ( 1 ) 5 x + C 2 6 ( 1 ) 4 ( x ) 2 - C 3 6 ( 1 ) 3 ( x ) 3 + C 4 6 ( 1 ) 2 ( x ) 4 - C 5 6 ( 1 ) ( x ) 5 + C 6 6 ( x ) 6 Now C 0 6 = 1 = C 6 6 C 1 6 = 6 = C 5 6 C 2 6 = 6 × 5 1 × 2 = 15 = C 4 6 C 3 6 = 6 × 5 × 4 1 × 2 × 3 = 20 ∴ ( a + b x ) ( 1 - x ) 6 = ( a + b x ) 1 - 6 x + 15 x 2 - 20 x 3 + 15 x 4 - 6 x 5 + x 6 = a + ( - 6 a + b ) x + ( 15 a - 6 b ) x 2 + ⋯ ( 1 ) It is given that, ( a + b x ) ( 1 - x ) 6 = 3 - 20 x + c x 2 + ⋯ ( 2 ) From ( 1 ) and ( 2 ) , we get a = 3 ⇒ - 6 a + b = - 20 . . . ( 3 ) ⇒ 15 a - 6 b = c . . . ( 4 ) Substituting a = 3 in 3 we get ⇒ - 6 ( 3 ) + b = - 20 ∴ b = - 2 Substituting a = 3 a n d b = - 2 i n ( 4 ) 15 ( 3 ) - 6 ( - 2 ) = c ⇒ 45 + 12 = c ∴ c = 57 Hence a = 3 , b = - 2 and c = 57

Suppose ( 1 + k x ) n = 1 - 12 x + 60 x 2 - … find k and n

Given that ( 1 + k x ) n = 1 - 12 x + 60 x 2 - ⋯ Applying binomial theorem for ( a + b ) n , we get ( a + b ) n = n C 0 · a n + C 1 n · a n - 1 · b + … … + C n - 1 n a · b n - 1 + C n n · b n where C r n = n ! r ! ( n - r ) ! = n · ( n - 1 ) · ( n - 2 ) · … · · r terms r ! ( 1 + k x ) n = C 0 n ( 1 ) n + C 1 n ( 1 ) n - 1 ( k x ) + C 2 n ( 1 ) n - 2 ( k x ) 2 + ⋯ ( 1 + k x ) n = 1 + n k x + n ( n - 1 ) 2 · k 2 x 2 + ⋯ ( 1 ) it is given that ( 1 + k x ) n = 1 - 12 x + 60 x 2 - ⋯ ( 2 ) From ( 1 ) and ( 2 ) , we get, n k = - 12 . . . ( 3 ) n ( n - 1 ) 2 · k 2 = 60 . . . ( 4 ) ⇒ n 2 - n k 2 = 120 ⇒ n 2 k 2 - n k 2 = 120 ⇒ ( n k ) 2 - ( n k ) k = 120 ⇒ ( - 12 ) 2 - ( - 12 k ) = 120 ⇒ 144 + 12 k = 120 ⇒ 12 k = - 24 ∴ k = - 2 from ( 3 ) , n ( - 2 ) = - 12 ∴ n = 6 Hence, n = 6 and k = - 2