Question Number 103513 by bemath last updated on 15/Jul/20

coefficient of x^5 in expansion (1+x^2 )^5 (1+x)^4 equal to (a) 40 (b) 45 (c) 50 (d) 55 (e) 60

Answered by OlafThorendsen last updated on 15/Jul/20

C_5 ^1 C_4 ^3 +C_5 ^2 C_4 ^1 = 5×4+((5!)/(3!2!))×4 = 60 ⇒ (e)

Commented bybemath last updated on 15/Jul/20

how do you got it ?

Commented byOlafThorendsen last updated on 15/Jul/20

(1+x^2 )^5 = Σ_(k=0) ^5 C_5 ^k x^(2k) (1+x)^4 = Σ_(k=0) ^4 C_4 ^k x^k To obtain x^5 two solutions only : x^2 ×x^3 or x^4 ×x^1 Then C_5 ^2 C_4 ^3 +C_5 ^4 C_4 ^1

Answered by Rio Michael last updated on 15/Jul/20

(1 + x^2 )^5 = Σ_(r=0) ^5 ^5 C_r x^(2r) ......(i) (1 + x)^4 = Σ_(r=0) ^4 ^4 C_r x^r ......(ii) ⇒ (1 + x^2 )^5 (1 + x)^4 = Σ_(r=0) ^5 ^5 C_r x^(2r) . Σ_(r=0) ^4 ^4 C_r x^r when r = 1 for (i) and r = 3 for (ii) we get x^5 ⇒ possibility is: ^5 C_1 .^4 C_3 also when r = 2 for (i) and r = 1 for (ii) we get x^5 ⇒ possibility is: ^5 C_2 .^4 C_1 ⇒ all possibilities are:^5 C_1 .^4 C_3 + ^5 C_2 .^4 C_1 an extended way of the solution above.