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 ■