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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 ?

$$\mathrm{If}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:{x}^{{k}} \:\mathrm{and}\:{x}^{{k}+\mathrm{1}} \:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{expansion}\:\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{19}} \:\mathrm{are}\:\mathrm{equal}\:,\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:? \\ $$

Commented by bobhans last updated on 27/Jun/20

thank you both

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$

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

$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{19}} =\Sigma{C}_{\mathrm{19}} ^{{k}} \left(\mathrm{3}{x}\right)^{{k}} \mathrm{2}^{\mathrm{19}−{k}} \\ $$$$\Rightarrow{C}_{\mathrm{19}} ^{{m}} \mathrm{3}^{{m}} \mathrm{2}^{\mathrm{19}−{m}} ={C}_{\mathrm{19}} ^{{m}+\mathrm{1}} \mathrm{3}^{{m}+\mathrm{1}} \mathrm{2}^{\mathrm{19}−{m}−\mathrm{1}} \Rightarrow \\ $$$$\frac{{C}_{\mathrm{19}} ^{{m}} }{{C}_{\mathrm{19}} ^{{m}+\mathrm{1}} }=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\left({m}+\mathrm{1}\right)!.\left(\mathrm{18}−{m}\right)!}{{m}!.\left(\mathrm{19}−{m}\right)!}=\frac{\mathrm{3}}{\mathrm{2}}\Leftrightarrow\mathrm{2}\left({m}+\mathrm{1}\right)=\mathrm{3}\left(\mathrm{19}−{m}\right) \\ $$$$\Rightarrow\mathrm{5}{m}=\mathrm{55}\Rightarrow{m}=\mathrm{11} \\ $$

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 ■

$$\mathrm{by}\:\mathrm{Binomial}\:\mathrm{Theorem} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{19}} \:=\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\mathrm{19}} {\sum}}\begin{pmatrix}{\mathrm{19}}\\{\:\:\mathrm{n}}\end{pmatrix}\:\left(\mathrm{2}\right)^{\mathrm{19}−\mathrm{n}} \:\left(\mathrm{3}{x}\right)^{{n}} \\ $$$${we}\:{want}\:{to}\:{find}\:{k}\:{so}\:{that}\: \\ $$$$\begin{pmatrix}{\mathrm{19}}\\{\:\:{k}}\end{pmatrix}\:\left(\mathrm{2}\right)^{\mathrm{19}−{k}} \:\left(\mathrm{3}{x}\right)^{{k}} \:=\:\begin{pmatrix}{\mathrm{19}}\\{{k}+\mathrm{1}}\end{pmatrix}\:\left(\mathrm{2}\right)^{\mathrm{18}−{k}} \:\left(\mathrm{3}{x}\right)^{{k}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}.\frac{\mathrm{19}!}{{k}!\:\left(\mathrm{19}−\mathrm{k}\right)!}\:=\:\mathrm{3}.\frac{\mathrm{19}!}{\left(\mathrm{k}+\mathrm{1}\right)!\:\left(\mathrm{18}−\mathrm{k}\right)!} \\ $$$$\mathrm{2}.\frac{\mathrm{19}!}{\left(\mathrm{19}−\mathrm{k}\right).\mathrm{k}!\:\left(\mathrm{18}−\mathrm{k}\right)!}\:=\:\mathrm{3}.\frac{\mathrm{19}!}{\left(\mathrm{k}+\mathrm{1}\right).\mathrm{k}!\:\left(\mathrm{18}−\mathrm{k}\right)!} \\ $$$$\frac{\mathrm{2}}{\mathrm{19}−\mathrm{k}}\:=\:\frac{\mathrm{3}}{\mathrm{k}+\mathrm{1}}\:\Leftrightarrow\:{k}\:=\:\mathrm{11}\:\blacksquare\: \\ $$

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