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Question Number 154225 by ZiYangLee last updated on 15/Sep/21

$$\mathrm{In}\mathrm{the}\mathrm{expansion}{(1+\frac{1}{2}x)}^n,\mathrm{the}\mathrm{coefficient}$$$$\mathrm{of}{x}^m\mathrm{is}\frac{5}{4}\mathrm{times}\mathrm{the}\mathrm{coefficient}\mathrm{of}{x}^{m+1}.$$$$a)\mathrm{Show}\mathrm{that}5n-13m=8$$$$b)\mathrm{If}m\mathrm{and}n\mathrm{are}\mathrm{positive}\mathrm{integers},\mathrm{such}\mathrm{that}$$$$m⩽n,\mathrm{determine}\mathrm{the}\mathrm{smallest}\mathrm{values}\mathrm{of}$$$$m\mathrm{and}n.$$

Commented by ZiYangLee last updated on 15/Sep/21

$$\mathrm{a})T_{m+1}={}^n{C}_m{\left(\frac{1}{2}x\right)}^m$$$$T_{(m+1)+1}={}^n{C}_{m+1}{\left(\frac{1}{2}x\right)}^{m+1}$$$$T_{m+1}=\frac{5}{4}{T}_{(m+1)+1}$$$${}^n{C}_m{\left(\frac{1}{2}\right)}^m=\frac{5}{4}{}^n{C}_{m+1}{\left(\frac{1}{2}\right)}^{m+1}$$$$\frac{n!}{m!(n-m)!}=\frac{5}{4}\times \frac{n!}{(m+1)!(n-m-1)!}\times \frac{1}{2}$$$$\frac{(n-m-1)!(m+1)!}{m!(n-m)!}=\frac{5}{8}$$$$\frac{(n-m-1)!(m+1)m!}{m!(n-m)(n-m-1)!}=\frac{5}{8}$$$$\frac{m+1}{n-m}=\frac{5}{8}$$$$8m+8=5n-5m$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by mr W last updated on 15/Sep/21

$${(1+\frac{x}{2})}^n=\underset{k=0}{\sum ^n}{C}_k^n\frac{x^k}{2^k}$$$$a_m=\frac{C_m^n}{2^m}$$$$a_{m+1}=\frac{C_{m+1}^n}{2^{m+1}}$$$$\frac{a_m}{a_{m+1}}=\frac{C_m^n}{2^m}\times \frac{2^{m+1}}{C_{m+1}^n}=\frac{2(m+1)!(n-m-1)!}{m!(n-m)!}$$$$=\frac{2(m+1)}{(n-m)}=\frac{5}{4}$$$$\frac{m+1}{n-m}=\frac{5}{8}$$$$\Rightarrow 5n-13m=8$$$$$$$$\Rightarrow n=13k-1withk⩾1$$$$\Rightarrow m=5k-1$$$$$$$$n_{min}=13-1=12$$$$m_{min}=5-1=4$$

Commented by ZiYangLee last updated on 15/Sep/21

$$wownice...!$$

Commented by Tawa11 last updated on 15/Sep/21

$$\mathrm{Weldone}\mathrm{sir}$$

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