Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 7102 by Tawakalitu. last updated on 10/Aug/16

Commented by 123456 last updated on 10/Aug/16

S_0 =2 S_1 =2

$${S}_{\mathrm{0}} =\mathrm{2} \\ $$$${S}_{\mathrm{1}} =\mathrm{2} \\ $$

Commented by prakash jain last updated on 10/Aug/16

x+y=2 xy=4 (x+y)^2 −4xy=(x−y)^2 =4−16=−12 x−y=2(√3)i x=1+i(√3)=2e^(iπ/3) y=1−i(√3)=2e^(−iπ/3) x^n +y^n =2^n e^(inπ/3) +2^n e^(−inπ/3) =2^(n+1) cos ((nπ)/3) S_(n+1) =2^(n+2) cos (((n+1)π)/3) S_(n−1) =2^n cos (((n−1)π)/3) S_(n+1) +4S_(n−1) =2^(n+2) [cos (((n+1)π)/3)+cos (((n−1)π)/3)] =2^(n+2) [2cos ((nπ)/3)cos (π/3)]=2^(n+3) cos ((nπ)/3)=4[2^(n+1) cos ((nπ)/3)] =2[2^(n+1) cos ((nπ)/3)×(1/2)]=2S_n

$${x}+{y}=\mathrm{2} \\ $$$${xy}=\mathrm{4} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}=\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4}−\mathrm{16}=−\mathrm{12} \\ $$$${x}−{y}=\mathrm{2}\sqrt{\mathrm{3}}{i} \\ $$$${x}=\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\pi/\mathrm{3}} \\ $$$${y}=\mathrm{1}−{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{−{i}\pi/\mathrm{3}} \\ $$$${x}^{{n}} +{y}^{{n}} =\mathrm{2}^{{n}} {e}^{{in}\pi/\mathrm{3}} +\mathrm{2}^{{n}} {e}^{−{in}\pi/\mathrm{3}} =\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}} \\ $$$${S}_{{n}+\mathrm{1}} =\mathrm{2}^{{n}+\mathrm{2}} \mathrm{cos}\:\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}} \\ $$$${S}_{{n}−\mathrm{1}} =\mathrm{2}^{{n}} \mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{3}} \\ $$$${S}_{{n}+\mathrm{1}} +\mathrm{4}{S}_{{n}−\mathrm{1}} =\mathrm{2}^{{n}+\mathrm{2}} \left[\mathrm{cos}\:\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}+\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{3}}\right] \\ $$$$=\mathrm{2}^{{n}+\mathrm{2}} \left[\mathrm{2cos}\:\frac{{n}\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\right]=\mathrm{2}^{{n}+\mathrm{3}} \mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}=\mathrm{4}\left[\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}\right] \\ $$$$=\mathrm{2}\left[\mathrm{2}^{{n}+\mathrm{1}} \mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}\right]=\mathrm{2}{S}_{{n}} \\ $$

Commented by Tawakalitu. last updated on 10/Aug/16

I really appreciate your effort, thank you so much

$${I}\:{really}\:{appreciate}\:{your}\:{effort},\:{thank}\:{you}\:{so}\:{much} \\ $$

Commented by Tawakalitu. last updated on 10/Aug/16

Workings please

$${Workings}\:{please} \\ $$

Commented by Tawakalitu. last updated on 10/Aug/16

please edit your wotkings and show how p = 4, q = 4. thanks

$${please}\:{edit}\:{your}\:{wotkings}\:{and}\:{show}\:{how}\:{p}\:=\:\mathrm{4},\:{q}\:=\:\mathrm{4}.\:{thanks}\: \\ $$

Commented by Yozzii last updated on 10/Aug/16

Check your second to last line of text. 2^(n+2) [2cos((nπ)/3)cos(π/3)]=2^(n+2) [2cos((nπ)/3)×(1/2)]=2×2^(n+1) cos((nπ)/3) ⇒p=2 ∴ pq=8

$${Check}\:{your}\:{second}\:{to}\:{last}\:{line}\:{of}\:{text}. \\ $$$$\mathrm{2}^{{n}+\mathrm{2}} \left[\mathrm{2}{cos}\frac{{n}\pi}{\mathrm{3}}{cos}\frac{\pi}{\mathrm{3}}\right]=\mathrm{2}^{{n}+\mathrm{2}} \left[\mathrm{2}{cos}\frac{{n}\pi}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}\right]=\mathrm{2}×\mathrm{2}^{{n}+\mathrm{1}} {cos}\frac{{n}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{p}=\mathrm{2}\:\therefore\:{pq}=\mathrm{8} \\ $$

Commented by prakash jain last updated on 10/Aug/16

Thanks. Corrected.

$$\mathrm{Thanks}.\:\mathrm{Corrected}. \\ $$

Commented by prakash jain last updated on 10/Aug/16

Please see last three lines for relation between S_(n+1) , S_(n−1) and S_n

$$\mathrm{Please}\:\mathrm{see}\:\mathrm{last}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{for}\:\mathrm{relation} \\ $$$$\mathrm{between}\:{S}_{{n}+\mathrm{1}} ,\:{S}_{{n}−\mathrm{1}} \:{and}\:{S}_{{n}} \\ $$

Commented by Tawakalitu. last updated on 10/Aug/16

Seen thanks so much, i can understand now

$${Seen}\:{thanks}\:{so}\:{much},\:{i}\:{can}\:{understand}\:{now} \\ $$

Answered by prakash jain last updated on 10/Aug/16

p=2 q=4 pq=16

$${p}=\mathrm{2} \\ $$$${q}=\mathrm{4} \\ $$$${pq}=\mathrm{16} \\ $$

Commented by Tawakalitu. last updated on 10/Aug/16

Workings please

$${Workings}\:{please}\: \\ $$

Commented by Yozzii last updated on 10/Aug/16

p=2, q=4, pq=8

$${p}=\mathrm{2},\:{q}=\mathrm{4},\:{pq}=\mathrm{8} \\ $$

Answered by nume1114 last updated on 12/Aug/16

{ ((x+y=2)),((xy=4)),((S_n =x^n +y^n )) :} S_(n+1) =x^(n+1) +y^(n+1) =(x+y)(x^n +y^n )−xy(x^(n−1) +y^(n−1) ) ((( (x+y)(x^n +y^n ))),((=x^(n+1) +y^(n+1) +x^n y+xy^n )),((=x^(n+1) +y^(n+1) +xy(x^(n−1) +y^(n−1) ))) ) =(x+y)S_n −xyS_(n−1) =2S_n −4S_(n−1) 2S_n =S_(n+1) +4S_(n−1) p=2 , q=4 pq=8

$$\begin{cases}{{x}+{y}=\mathrm{2}}\\{{xy}=\mathrm{4}}\\{{S}_{{n}} ={x}^{{n}} +{y}^{{n}} }\end{cases} \\ $$$${S}_{{n}+\mathrm{1}} ={x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({x}+{y}\right)\left({x}^{{n}} +{y}^{{n}} \right)−{xy}\left({x}^{{n}−\mathrm{1}} +{y}^{{n}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\begin{pmatrix}{\:\:\:\:\left({x}+{y}\right)\left({x}^{{n}} +{y}^{{n}} \right)}\\{={x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{x}^{{n}} {y}+{xy}^{{n}} }\\{={x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{xy}\left({x}^{{n}−\mathrm{1}} +{y}^{{n}−\mathrm{1}} \right)}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({x}+{y}\right){S}_{{n}} −{xyS}_{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{S}_{{n}} −\mathrm{4}{S}_{{n}−\mathrm{1}} \\ $$$$\:\:\mathrm{2}{S}_{{n}} ={S}_{{n}+\mathrm{1}} +\mathrm{4}{S}_{{n}−\mathrm{1}} \\ $$$${p}=\mathrm{2}\:,\:\:{q}=\mathrm{4} \\ $$$${pq}=\mathrm{8} \\ $$

Commented by Rasheed Soomro last updated on 12/Aug/16

Very Nice!

$${Very}\:{Nice}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com