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Question Number 114208 by aurpeyz last updated on 17/Sep/20

find the greatest coeeficient in the expansion  of (3+4x)^(−5)

$${find}\:{the}\:{greatest}\:{coeeficient}\:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} \\ $$

Commented by I want to learn more last updated on 17/Sep/20

Please how sir

$$\mathrm{Please}\:\mathrm{how}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 17/Sep/20

no maximum, no minimum!

$${no}\:{maximum},\:{no}\:{minimum}! \\ $$

Commented by aurpeyz last updated on 18/Sep/20

why?

$${why}? \\ $$

Commented by mr W last updated on 18/Sep/20

(3+4x)^(−5)   =3^(−5) ×(1+(4/3)x)^(−5)   =3^(−5) Σ_(k=0) ^∞ (−(4/3))^k C_4 ^(k+4) x^k   with increasing k, ((4/3))^k is increasing  and C_4 ^(k+4)  is aslo increasing. therefore  there is no max/min for 3^(−5) (−(4/3))^k C_4 ^(k+4) ,  it →+∞ or −∞.

$$\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} ×\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}{x}\right)^{−\mathrm{5}} \\ $$$$=\mathrm{3}^{−\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{4}}{\mathrm{3}}\right)^{{k}} {C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}} \\ $$$${with}\:{increasing}\:{k},\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{k}} {is}\:{increasing} \\ $$$${and}\:{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \:{is}\:{aslo}\:{increasing}.\:{therefore} \\ $$$${there}\:{is}\:{no}\:{max}/{min}\:{for}\:\mathrm{3}^{−\mathrm{5}} \left(−\frac{\mathrm{4}}{\mathrm{3}}\right)^{{k}} {C}_{\mathrm{4}} ^{{k}+\mathrm{4}} , \\ $$$${it}\:\rightarrow+\infty\:{or}\:−\infty. \\ $$

Commented by aurpeyz last updated on 18/Sep/20

i think there is more to this topic than i know

$${i}\:{think}\:{there}\:{is}\:{more}\:{to}\:{this}\:{topic}\:{than}\:{i}\:{know} \\ $$

Commented by mr W last updated on 18/Sep/20

with that what you know you can  solve alot of questions, such as this  one. just apply!

$${with}\:{that}\:{what}\:{you}\:{know}\:{you}\:{can} \\ $$$${solve}\:{alot}\:{of}\:{questions},\:{such}\:{as}\:{this} \\ $$$${one}.\:{just}\:{apply}! \\ $$

Commented by aurpeyz last updated on 18/Sep/20

thanks alot Sir

$${thanks}\:{alot}\:{Sir} \\ $$

Commented by aurpeyz last updated on 18/Sep/20

when i try to find the greatest coeeficient  using Σ_(r=0) ^∞ C_r ^(5+r−1) ((4/3))^r (−1)^(r )  I arrived at a negative   value for r. I think it is a proof that there is no   greatest coefficient.

$${when}\:{i}\:{try}\:{to}\:{find}\:{the}\:{greatest}\:{coeeficient} \\ $$$${using}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{r}} ^{\mathrm{5}+{r}−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{r}} \left(−\mathrm{1}\right)^{{r}\:} \:\mathrm{I}\:\mathrm{arrived}\:\mathrm{at}\:\mathrm{a}\:\mathrm{negative}\: \\ $$$$\mathrm{value}\:\mathrm{for}\:\mathrm{r}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{proof}\:\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\: \\ $$$$\mathrm{greatest}\:\mathrm{coefficient}. \\ $$

Commented by mr W last updated on 18/Sep/20

no proof needed. it′s clear:  C_4 ^(4+r) ((4/3))^r  is always increasing and  (−1)^(r ) =1 for even r and −1 for odd r.  so there is neither maximum nor  minimum coefficient.

$${no}\:{proof}\:{needed}.\:{it}'{s}\:{clear}: \\ $$$${C}_{\mathrm{4}} ^{\mathrm{4}+{r}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{r}} \:{is}\:{always}\:{increasing}\:{and} \\ $$$$\left(−\mathrm{1}\right)^{{r}\:} =\mathrm{1}\:{for}\:{even}\:{r}\:{and}\:−\mathrm{1}\:{for}\:{odd}\:{r}. \\ $$$${so}\:{there}\:{is}\:{neither}\:{maximum}\:{nor} \\ $$$${minimum}\:{coefficient}. \\ $$

Commented by mr W last updated on 18/Sep/20

it′s something like  1,−2,3,−4,5,−6,....  it has no maximum and no minimum.

$${it}'{s}\:{something}\:{like} \\ $$$$\mathrm{1},−\mathrm{2},\mathrm{3},−\mathrm{4},\mathrm{5},−\mathrm{6},.... \\ $$$${it}\:{has}\:{no}\:{maximum}\:{and}\:{no}\:{minimum}. \\ $$

Commented by aurpeyz last updated on 18/Sep/20

anytime i have a question. how can i know  without proof that there is no max or min.  did you determine it by ((4/3))^r  or the nature  of the question being (3+4x)^(−5) .    i have not seen any textbook that deals with  greatest coefficient of question like that.   i thought it is because of the positive sign (+)   between the binomial.    what if we had a question with ((3/4))^x  ?

$${anytime}\:{i}\:{have}\:{a}\:{question}.\:{how}\:{can}\:{i}\:{know} \\ $$$${without}\:{proof}\:{that}\:{there}\:{is}\:{no}\:{max}\:{or}\:{min}. \\ $$$${did}\:{you}\:{determine}\:{it}\:{by}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{r}} \:{or}\:{the}\:{nature} \\ $$$${of}\:{the}\:{question}\:{being}\:\left(\mathrm{3}+\mathrm{4}{x}\right)^{−\mathrm{5}} . \\ $$$$ \\ $$$${i}\:{have}\:{not}\:{seen}\:{any}\:{textbook}\:{that}\:{deals}\:{with} \\ $$$${greatest}\:{coefficient}\:{of}\:{question}\:{like}\:{that}.\: \\ $$$${i}\:{thought}\:{it}\:{is}\:{because}\:{of}\:{the}\:{positive}\:{sign}\:\left(+\right)\: \\ $$$${between}\:{the}\:{binomial}. \\ $$$$ \\ $$$${what}\:{if}\:{we}\:{had}\:{a}\:{question}\:{with}\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} \:? \\ $$

Commented by mr W last updated on 18/Sep/20

(−1)^r C_4 ^(4+r) ((4/3))^r has no max and no min.  (−1)^r C_4 ^(4+r) ((3/4))^r  has max and min.

$$\left(−\mathrm{1}\right)^{{r}} {C}_{\mathrm{4}} ^{\mathrm{4}+{r}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{r}} {has}\:{no}\:{max}\:{and}\:{no}\:{min}. \\ $$$$\left(−\mathrm{1}\right)^{{r}} {C}_{\mathrm{4}} ^{\mathrm{4}+{r}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{r}} \:{has}\:{max}\:{and}\:{min}. \\ $$

Commented by aurpeyz last updated on 18/Sep/20

thank you Sir. i will attempt it

$${thank}\:{you}\:{Sir}.\:{i}\:{will}\:{attempt}\:{it} \\ $$

Commented by aurpeyz last updated on 18/Sep/20

lets say i have (4+3x)^(−5)   (4+3x)^(−5) =4^(−5) [(−1)^r Σ_(r=0) ^∞ C_r ^(4+r) ((3/4))^r ]  (−1)^r C_r ^(4+r) ((3/4))^r ≥ (−1)^(r+1) C_(r+1) ^(5+r) ((3/4))  C_4 ^(4+r) ≥(−1)C_4 ^(5+r) ((3/4))^1   r=((−19)/7).   the negativity gives me issue.    is it that any question like (4+3x)^(−5)  [with   positive sign in between binomial of negative  powers] usually dont have max or min?

$${lets}\:{say}\:{i}\:{have}\:\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} \\ $$$$\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} =\mathrm{4}^{−\mathrm{5}} \left[\left(−\mathrm{1}\right)^{{r}} \underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{r}} ^{\mathrm{4}+{r}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{r}} \right] \\ $$$$\left(−\mathrm{1}\right)^{{r}} {C}_{{r}} ^{\mathrm{4}+{r}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{r}} \geqslant\:\left(−\mathrm{1}\right)^{{r}+\mathrm{1}} {C}_{{r}+\mathrm{1}} ^{\mathrm{5}+{r}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${C}_{\mathrm{4}} ^{\mathrm{4}+{r}} \geqslant\left(−\mathrm{1}\right){C}_{\mathrm{4}} ^{\mathrm{5}+{r}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}} \\ $$$${r}=\frac{−\mathrm{19}}{\mathrm{7}}.\: \\ $$$${the}\:{negativity}\:{gives}\:{me}\:{issue}. \\ $$$$ \\ $$$${is}\:{it}\:{that}\:{any}\:{question}\:{like}\:\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} \:\left[{with}\:\right. \\ $$$${positive}\:{sign}\:{in}\:{between}\:{binomial}\:{of}\:{negative} \\ $$$$\left.{powers}\right]\:{usually}\:{dont}\:{have}\:{max}\:{or}\:{min}? \\ $$

Commented by mr W last updated on 18/Sep/20

to treat the positive coefficients  you should only see the even terms,  i.e. r=2k  ......

$${to}\:{treat}\:{the}\:{positive}\:{coefficients} \\ $$$${you}\:{should}\:{only}\:{see}\:{the}\:{even}\:{terms}, \\ $$$${i}.{e}.\:{r}=\mathrm{2}{k} \\ $$$$...... \\ $$

Commented by aurpeyz last updated on 19/Sep/20

please shed little light on this. r=2k? how  is this done?

$${please}\:{shed}\:{little}\:{light}\:{on}\:{this}.\:{r}=\mathrm{2}{k}?\:{how} \\ $$$${is}\:{this}\:{done}? \\ $$

Commented by mr W last updated on 19/Sep/20

even terms (r=2k) have positive   coefficients, odd terms (r=2k+1)  have negative coefficients. just think!

$${even}\:{terms}\:\left({r}=\mathrm{2}{k}\right)\:{have}\:{positive}\: \\ $$$${coefficients},\:{odd}\:{terms}\:\left({r}=\mathrm{2}{k}+\mathrm{1}\right) \\ $$$${have}\:{negative}\:{coefficients}.\:{just}\:{think}! \\ $$

Commented by aurpeyz last updated on 20/Sep/20

okay

$${okay}\: \\ $$

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