Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 178937 by Tawa11 last updated on 22/Oct/22

Find the greatest coefficient in expansion of:    (6   −   4x)^(−  3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{expansion}\:\mathrm{of}:\:\:\:\:\left(\mathrm{6}\:\:\:−\:\:\:\mathrm{4x}\right)^{−\:\:\mathrm{3}} \\ $$

Answered by mr W last updated on 23/Oct/22

(6−4x)^(−3)   =6^(−3) (1−((2x)/3))^(−3)   =6^(−3) Σ_(k=0) ^∞ C_2 ^(k+2) ((2/3))^k x^k   a_k =6^(−3) ((2/3))^k C_2 ^(k+2) =((2^(k−4) (k+2)(k+1))/3^(k+3) )  a_(k+1) =((2^(k−3) (k+3)(k+2))/3^(k+4) )  a_k >a_(k+1)   ((2^(k−4) (k+2)(k+1))/3^(k+3) )>((2^(k−3) (k+3)(k+2))/3^(k+4) )  (((k+1))/1)>((2(k+3))/3)  ⇒k>3  ⇒a_4  is maximum!  a_(max) =a_4 =((2^(4−4) (4+2)(4+1))/3^(4+3) )=((10)/(729)) ✓

$$\left(\mathrm{6}−\mathrm{4}{x}\right)^{−\mathrm{3}} \\ $$$$=\mathrm{6}^{−\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{−\mathrm{3}} \\ $$$$=\mathrm{6}^{−\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{k}} {x}^{{k}} \\ $$$${a}_{{k}} =\mathrm{6}^{−\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{k}} {C}_{\mathrm{2}} ^{{k}+\mathrm{2}} =\frac{\mathrm{2}^{{k}−\mathrm{4}} \left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right)}{\mathrm{3}^{{k}+\mathrm{3}} } \\ $$$${a}_{{k}+\mathrm{1}} =\frac{\mathrm{2}^{{k}−\mathrm{3}} \left({k}+\mathrm{3}\right)\left({k}+\mathrm{2}\right)}{\mathrm{3}^{{k}+\mathrm{4}} } \\ $$$${a}_{{k}} >{a}_{{k}+\mathrm{1}} \\ $$$$\frac{\mathrm{2}^{{k}−\mathrm{4}} \left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right)}{\mathrm{3}^{{k}+\mathrm{3}} }>\frac{\mathrm{2}^{{k}−\mathrm{3}} \left({k}+\mathrm{3}\right)\left({k}+\mathrm{2}\right)}{\mathrm{3}^{{k}+\mathrm{4}} } \\ $$$$\frac{\left({k}+\mathrm{1}\right)}{\mathrm{1}}>\frac{\mathrm{2}\left({k}+\mathrm{3}\right)}{\mathrm{3}} \\ $$$$\Rightarrow{k}>\mathrm{3} \\ $$$$\Rightarrow{a}_{\mathrm{4}} \:{is}\:{maximum}! \\ $$$${a}_{{max}} ={a}_{\mathrm{4}} =\frac{\mathrm{2}^{\mathrm{4}−\mathrm{4}} \left(\mathrm{4}+\mathrm{2}\right)\left(\mathrm{4}+\mathrm{1}\right)}{\mathrm{3}^{\mathrm{4}+\mathrm{3}} }=\frac{\mathrm{10}}{\mathrm{729}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 23/Oct/22

Great, God bless you sir. I appreciare.

$$\mathrm{Great},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciare}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com