Let α , β , γ and δ be the coefficients of x 7 , x 5 , x 3 and x respectively in the expansion of ( x + x 3 − 1 ) 5 + ( x − x 3 − 1 ) 5 , x 1 . If u and v satisfy the equations αu + β v = 18 γ

Question

Let α , β , γ and δ be the coefficients of x 7 , x 5 , x 3 and x respectively in the expansion of ( x + x 3 − 1 ​ ) 5 + ( x − x 3 − 1 ​ ) 5 , x > 1 . If u and v satisfy the equations ​ αu + β v = 18 γ u + δ v = 20 ​ then u + v equals :

TestHub · Accepted Answer

​ ( x + x 3 − 1 ​ ) 5 + ( x − x 3 − 1 ​ ) 5 − [ 5 C 0 ​ x 5 + 5 C 1 ​ x 4 ( x 3 − 1 ​ ) + …… + 5 C 5 ​ ( x 3 − 1 ​ ) 5 ] + [ 5 C 0 ​ x 5 − 5 C 1 ​ x 4 ( x 3 − 1 ​ ) + …… . + − 5 C 5 ​ ( x 3 − 1 ​ ) 5 ] = 2 [ x 5 + 5 C 2 ​ x 3 ( x 3 − 1 ) + 5 C 4 ​ ( x 3 − 1 ) 2 ] = 10 x 7 + 20 x 6 + 2 x 5 − 20 x 4 − 20 x 3 + 10 x ​ Now α = 10 , β = 2 , γ = − 20 , δ = 10 Also, ​ 10 u + 2 v = 18 − 20 u + 10 v = 20 ​ } u = 1 , v = 4 u + v = 5 ​

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