Question Number 100587 by bobhans last updated on 27/Jun/20

If the coefficients of x^k  and x^(k+1)  in the   expansion (2+3x)^(19)  are equal , what is  the value of k ?

Commented bybobhans last updated on 27/Jun/20

thank you both

Answered by maths mind last updated on 27/Jun/20

(2+3x)^(19) =ΣC_(19) ^k (3x)^k 2^(19−k)   ⇒C_(19) ^m 3^m 2^(19−m) =C_(19) ^(m+1) 3^(m+1) 2^(19−m−1) ⇒  (C_(19) ^m /C_(19) ^(m+1) )=(3/2)  ⇒(((m+1)!.(18−m)!)/(m!.(19−m)!))=(3/2)⇔2(m+1)=3(19−m)  ⇒5m=55⇒m=11

Answered by john santu last updated on 27/Jun/20

by Binomial Theorem  (2+3x)^(19)  = Σ_(n = 0) ^(19)  (((19)),((  n)) ) (2)^(19−n)  (3x)^n   we want to find k so that    (((19)),((  k)) ) (2)^(19−k)  (3x)^k  =  (((19)),((k+1)) ) (2)^(18−k)  (3x)^(k+1)   ⇒2.((19!)/(k! (19−k)!)) = 3.((19!)/((k+1)! (18−k)!))  2.((19!)/((19−k).k! (18−k)!)) = 3.((19!)/((k+1).k! (18−k)!))  (2/(19−k)) = (3/(k+1)) ⇔ k = 11 ■