Menu Close

find-the-greatest-coeeficient-of-1-x-2-2-i-have-applied-Q-125697-but-i-got-two-negative-values-of-n-after-i-have-made-k-2n-for-positive-coefficient-I-believe-it-should-have-a-greatest-coeefi

Tinku Tara 0 Comments
Pin




Question Number 125881 by aurpeyz last updated on 14/Dec/20
find the greatest coeeficient of (1+(x/2))^(−2) i have applied Q.125697 but i got two negative values of n after i have made k=2n for positive coefficient. I believe it should have a greatest coeeficient since ((1/2))^k decreases as as k increases thanks what do i do?
$${find}\:{the}\:{greatest}\:{coeeficient}\:{of} \\ $$$$\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)^{−\mathrm{2}} \\ $$$${i}\:{have}\:{applied}\:{Q}.\mathrm{125697}\:{but}\:{i}\:{got}\:{two} \\ $$$${negative}\:{values}\:{of}\:{n}\:{after}\:{i}\:{have} \\ $$$${made}\:{k}=\mathrm{2}{n}\:{for}\:{positive}\:{coefficient}. \\ $$$$ \\ $$$${I}\:{believe}\:{it}\:{should}\:{have}\:{a}\:{greatest} \\ $$$${coeeficient}\:{since}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} {decreases}\:{as} \\ $$$${as}\:{k}\:{increases} \\ $$$${thanks} \\ $$$${what}\:{do}\:{i}\:{do}?\: \\ $$
Commented by mr W last updated on 14/Dec/20
yes, there is a maximum (as well as minimum) coefficient. (1+(x/2))^(−2) =Σ_(k=0) ^∞ (−1)^k C_1 ^(k+1) (1/2^k )x^k a_(2n) =(1/2^(2n) )C_1 ^(2n+1) =((2n+1)/2^(2n) ) a_(2n+1) =−(1/2^(2n+1) )C_1 ^(2n+2) =−((n+1)/2^(2n) ) ((2n+1)/2^(2n) )>((2(n+1)+1)/2^(2(n+1)) ) 4(2n+1)>2n+3 n>−(1/6) ⇒n_(min) =0 ⇒a_0 is the largest coefficient ⇒a_0 =1
$${yes},\:{there}\:{is}\:{a}\:{maximum}\:\left({as}\:{well}\:{as}\right. \\ $$$$\left.{minimum}\right)\:{coefficient}. \\ $$$$\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)^{−\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{\mathrm{1}} ^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }{x}^{{k}} \\ $$$${a}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} } \\ $$$${a}_{\mathrm{2}{n}+\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{C}_{\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{2}} =−\frac{{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} } \\ $$$$ \\ $$$$\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }>\frac{\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}}{\mathrm{2}^{\mathrm{2}\left({n}+\mathrm{1}\right)} } \\ $$$$\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)>\mathrm{2}{n}+\mathrm{3} \\ $$$${n}>−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{n}_{{min}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{0}} \:{is}\:{the}\:{largest}\:{coefficient} \\ $$$$\Rightarrow{a}_{\mathrm{0}} =\mathrm{1} \\ $$
Commented by aurpeyz last updated on 15/Dec/20
wow. thats nice. thanks alot. i can now solve all problems on it
$${wow}.\:{thats}\:{nice}.\:{thanks}\:{alot}.\: \\ $$$${i}\:{can}\:{now}\:{solve}\:{all}\:{problems}\:{on}\:{it} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *