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.