Question

12. Find (r + 1)° + (x - 1)°. Hence or otherwise evaluate (V2 + )2

Answer 1

The given expression ( x+1 ) 6 − ( x−1 ) 6 and we have to evaluate ( 2 + 1 ) 6 − ( 2 − 1 ) 6 .

The formula for binomial expansion is ,

( a+b ) n = C n 0 a n + C n 1 a n−1 b+ C n 2 a n−2 b 2 +..........+ C n n−1 a. b n−1 + C n n b n

The expression ( x+1 ) 6 − ( x−1 ) 6 , can be expanded as

According to the question n=6 .

( x+1 ) 6 = C 6 0 x 6 1 0 + C 6 1 x 6−1 1 1 + C 6 2 x 6−2 1 2 + C 6 3 a 6−3 1 3 + C 6 4 x 6−4 1 4 + C 6 5 x 6−5 1 5 + C 6 6 x 6−6 1 6 = C 6 0 x 6 + C 6 1 x 5 + C 6 2 x 4 + C 6 3 x 3 + C 6 4 x 2 + C 6 5 x 1 + C 6 6 x 0 = x 6 +6 x 5 +15 x 4 +20 x 3 +15 x 2 +6x+1 (1)

( x−1 ) 6 = C 6 0 x 6−0 1 0 − C 6 1 x 6−1 1 1 + C 6 2 x 6−2 1 2 − C 6 3 x 6−3 1 3 + C 6 4 x 6−4 1 4 − C 6 5 x 6−5 1 5 + C 6 6 x 6−6 1 6 = C 6 0 x 6 − C 6 1 x 5 + C 6 2 x 4 − C 6 3 x 3 + C 6 4 x 2 − C 6 5 x 1 + C 6 6 x 0 = x 6 −6 x 5 +15 x 4 −20 x 3 +15 x 2 −6x+1 (2)

Subtract equation (2) from (1),

( x+1 ) 6 + ( x−1 ) 6 =[ [ C 6 0 x 6 1 0 + C 6 1 x 5 1 1 + C 6 2 x 4 1 2 + C 6 3 x 3 1 3 + C 6 4 x 2 1 4 + C 6 5 x 1 1 5 + C 6 6 x 0 1 6 ] −[ C 6 0 x 6 1 0 − C 6 1 x 5 1 1 + C 6 2 x 4 1 2 − C 6 3 x 3 1 3 + C 6 4 x 2 1 4 − C 6 5 x 1 1 5 + C 6 6 x 0 1 6 ] ] =2( C 6 0 x 6 + C 6 2 x 4 + 6 C 4 x 2 + C 6 6 ) =2( x 6 +15 x 4 +15 x 2 +1 ) (3)

Substitute the values of x= 2 in equation (3), we get

( 2 + 1 ) 6 − ( 2 − 1 ) 6 =2[ ( 2 ) 6 +15 ( 2 ) 4 +15 ( 2 ) 2 +1 ] =2( 8+15×4+15×2+1 ) =2( 8+60+30+1 ) =2( 99 ) =198

Thus the evaluation of the expression ( 2 + 1 ) 6 − ( 2 − 1 ) 6 is 198 .