Question Number 75242 by peter frank last updated on 08/Dec/19
$${Find}\:{th}\:{greatest}\:{coefficients} \\ $$$${in}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{3}{a}+\mathrm{5}{b}\right)^{\mathrm{18}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right){If}\:{three}\:{consecutive}\: \\ $$$${coefficient}\:{of}\:\left(\mathrm{1}+{x}\right)^{{n}} \:{are}\:\mathrm{28},\mathrm{56},\mathrm{70}. \\ $$$${find}\:{the}\:{value}\:{of}\:{n} \\ $$$$ \\ $$
Answered by mr W last updated on 08/Dec/19
$$\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{3}{a}+\mathrm{5}{b}\right)^{\mathrm{18}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{18}} {\sum}}{C}_{{k}} ^{\mathrm{18}} \left(\mathrm{3}{a}\right)^{{k}} \left(\mathrm{5}{b}\right)^{\mathrm{18}−{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{18}} {\sum}}\mathrm{3}^{{k}} \mathrm{5}^{\mathrm{18}−{k}} {C}_{{k}} ^{\mathrm{18}} {a}^{{k}} {b}^{\mathrm{18}−{k}} \\ $$$${a}_{{k}} =\mathrm{3}^{{k}} \mathrm{5}^{\mathrm{18}−{k}} {C}_{{k}} ^{\mathrm{18}} \\ $$$${a}_{{k}+\mathrm{1}} =\mathrm{3}^{{k}+\mathrm{1}} \mathrm{5}^{\mathrm{17}−{k}} {C}_{{k}+\mathrm{1}} ^{\mathrm{18}} \\ $$$$\frac{{a}_{{k}+\mathrm{1}} }{{a}_{{k}} }=\frac{\mathrm{3}^{{k}+\mathrm{1}} \mathrm{5}^{\mathrm{17}−{k}} {C}_{{k}+\mathrm{1}} ^{\mathrm{18}} }{\mathrm{3}^{{k}} \mathrm{5}^{\mathrm{18}−{k}} {C}_{{k}} ^{\mathrm{18}} }=\frac{\mathrm{3}\left(\mathrm{18}−{k}\right)}{\mathrm{5}\left({k}+\mathrm{1}\right)}<\mathrm{1} \\ $$$$\mathrm{5}{k}+\mathrm{5}>\mathrm{54}−\mathrm{3}{k} \\ $$$${k}>\frac{\mathrm{49}}{\mathrm{8}}=\mathrm{6}.\mathrm{12}=\mathrm{7}\:\Rightarrow{a}_{\mathrm{8}} <{a}_{\mathrm{7}} \\ $$$${max}.\:{a}_{{k}} \:{is}\:{a}_{\mathrm{7}} =\mathrm{3}^{\mathrm{7}} \mathrm{5}^{\mathrm{11}} {C}_{\mathrm{7}} ^{\mathrm{18}} =\mathrm{3398392968750000} \\ $$
Commented by peter frank last updated on 09/Dec/19
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by mr W last updated on 08/Dec/19
$$\left(\mathrm{2}\right) \\ $$$${C}_{{k}} ^{{n}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!}=\mathrm{28} \\ $$$${C}_{{k}+\mathrm{1}} ^{{n}} =\frac{{n}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}−\mathrm{1}\right)!}=\mathrm{56} \\ $$$$\frac{{n}−{k}}{{k}+\mathrm{1}}=\frac{\mathrm{56}}{\mathrm{28}}=\mathrm{2} \\ $$$$\Rightarrow{n}=\mathrm{3}{k}+\mathrm{2} \\ $$$${C}_{{k}+\mathrm{2}} ^{{n}} =\frac{{n}!}{\left({k}+\mathrm{2}\right)!\left({n}−{k}−\mathrm{2}\right)!}=\mathrm{70} \\ $$$$\frac{{n}−{k}−\mathrm{1}}{{k}+\mathrm{2}}=\frac{\mathrm{70}}{\mathrm{56}} \\ $$$$\Rightarrow\mathrm{28}{n}=\mathrm{63}{k}+\mathrm{98} \\ $$$$\Rightarrow\mathrm{28}\left(\mathrm{3}{k}+\mathrm{2}\right)=\mathrm{63}{k}+\mathrm{98} \\ $$$$\Rightarrow{k}=\mathrm{2} \\ $$$$\Rightarrow{n}=\mathrm{8} \\ $$
Commented by peter frank last updated on 09/Dec/19
$${thank}\:{you}\:{very}\:{much} \\ $$