Question and Answers Forum

AllQuestion and Answers: Page 1149

Question Number 100649 Answers: 0 Comments: 0

calculate Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) and Σ_(n=0) ^∞ (((−1)^n )/(4n+1))

$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{3n}+\mathrm{1}}\:\mathrm{and}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}} \\ $$

Question Number 100644 Answers: 3 Comments: 1

Question Number 100640 Answers: 2 Comments: 0

let( U_n ) be a sequence definied by: { ((U_0 =1)),((U_(n+1) =((3U_n +2)/(U_n +2)))) :} show that 0<U_n <2

$${let}\left(\:\boldsymbol{{U}}_{{n}} \right)\:{be}\:{a}\:{sequence}\:{definied}\:{by}: \\ $$$$\begin{cases}{\boldsymbol{{U}}_{\mathrm{0}} =\mathrm{1}}\\{\boldsymbol{{U}}_{{n}+\mathrm{1}} =\frac{\mathrm{3}\boldsymbol{{U}}_{\boldsymbol{{n}}} +\mathrm{2}}{\boldsymbol{{U}}_{\boldsymbol{{n}}} +\mathrm{2}}}\end{cases} \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\mathrm{0}<\boldsymbol{{U}}_{\boldsymbol{{n}}} <\mathrm{2} \\ $$

Question Number 100629 Answers: 1 Comments: 1

Question Number 100624 Answers: 1 Comments: 3

Find the value of (√(2+(√(2+(√(2+(√2)))))))...∞ using cos function

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}...\infty\:\mathrm{using}\:\mathrm{cos}\:\mathrm{function} \\ $$

Question Number 100622 Answers: 0 Comments: 2

Question Number 100618 Answers: 1 Comments: 0

Σ_(k=1) ^(k=n) ((ln(k))/2^k ) =?

$$\:\:\:\:\:\:\:\:\:\:\:\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{k}}=\boldsymbol{{n}}} {\sum}}\:\frac{\boldsymbol{{ln}}\left(\boldsymbol{{k}}\right)}{\mathrm{2}^{\boldsymbol{{k}}} }\:=? \\ $$

Question Number 100614 Answers: 0 Comments: 0

Σ_(k=0) ^(k=n−1) ((ln(k!))/2^(k+1) ) =? Any help ?

$$\:\:\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{k}}=\boldsymbol{{n}}−\mathrm{1}} {\sum}}\frac{\boldsymbol{{ln}}\left(\boldsymbol{{k}}!\right)}{\mathrm{2}^{\boldsymbol{{k}}+\mathrm{1}} }\:=?\:\:\:\: \\ $$$$\:\:\boldsymbol{\mathrm{A}{ny}}\:\boldsymbol{{help}}\:? \\ $$

Question Number 100613 Answers: 0 Comments: 0

Question Number 100606 Answers: 0 Comments: 0

∫e^(ix^(ix...∞) ) dx

$$\int{e}^{{ix}^{{ix}...\infty} } {dx} \\ $$

Question Number 100597 Answers: 2 Comments: 1

Question Number 100594 Answers: 2 Comments: 0

solve the differential equations 1- xcos (ln (x/y))dy−ydx=0 2- ydx+2xdy =2y((√x)/(cos^2 (y)))dy y(0)=π

$${solve}\:\:{the}\:{differential}\:\:{equations} \\ $$$$\mathrm{1}-\:\:{x}\mathrm{cos}\:\left(\mathrm{ln}\:\frac{{x}}{{y}}\right){dy}−{ydx}=\mathrm{0} \\ $$$$\mathrm{2}-\:\:{ydx}+\mathrm{2}{xdy}\:=\mathrm{2}{y}\frac{\sqrt{{x}}}{{cos}^{\mathrm{2}} \left({y}\right)}{dy}\:\:\:\:\:{y}\left(\mathrm{0}\right)=\pi \\ $$

Question Number 100590 Answers: 2 Comments: 0

∫_0 ^∞ (dx/((1+x^(18) )^2 ))

$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{18}} \right)^{\mathrm{2}} } \\ $$

Question Number 100587 Answers: 2 Comments: 1

If the coefficients of x^k and x^(k+1) in the expansion (2+3x)^(19) are equal , what is the value of k ?

$$\mathrm{If}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:{x}^{{k}} \:\mathrm{and}\:{x}^{{k}+\mathrm{1}} \:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{expansion}\:\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{19}} \:\mathrm{are}\:\mathrm{equal}\:,\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:? \\ $$

Question Number 100585 Answers: 0 Comments: 0

Given that G = {1,(x + yi),(x−yi)} form a group under complex multiplication, describe the locus of the point (x,y)

$$\:\mathrm{Given}\:\mathrm{that}\:\:{G}\:=\:\left\{\mathrm{1},\left({x}\:+\:{yi}\right),\left({x}−{yi}\right)\right\}\:\mathrm{form}\:\mathrm{a}\:\mathrm{group} \\ $$$$\mathrm{under}\:\mathrm{complex}\:\mathrm{multiplication},\:\mathrm{describe}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\left({x},{y}\right) \\ $$

Question Number 100584 Answers: 1 Comments: 0

∫i^i^(i......∞) dx

$$\int{i}^{{i}^{{i}......\infty} } {dx} \\ $$

Question Number 100583 Answers: 0 Comments: 0

A transformation f on a complex plane is defined by z′ = (1 +i)z −3 + 4i show that f is a simultitude with radius r and centre Ω to be determined. Determine to the invariant point under f.

$$\:\mathrm{A}\:\mathrm{transformation}\:{f}\:\mathrm{on}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{plane} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:{z}'\:=\:\left(\mathrm{1}\:+{i}\right){z}\:−\mathrm{3}\:+\:\mathrm{4}{i} \\ $$$$\:\mathrm{show}\:\mathrm{that}\:{f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{simultitude}\:\mathrm{with}\:\mathrm{radius}\:{r}\:\mathrm{and}\:\mathrm{centre} \\ $$$$\Omega\:\mathrm{to}\:\mathrm{be}\:\mathrm{determined}. \\ $$$$\mathrm{Determine}\:\mathrm{to}\:\mathrm{the}\:\mathrm{invariant}\:\mathrm{point}\:\mathrm{under}\:{f}. \\ $$

Question Number 100581 Answers: 0 Comments: 0

If α=((2π)/7) then prove that tanαtan2α+tan2αtan4α+tan4αtanα=−7

$${If}\:\alpha=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$${then}\:{prove}\:{that} \\ $$$${tan}\alpha{tan}\mathrm{2}\alpha+{tan}\mathrm{2}\alpha{tan}\mathrm{4}\alpha+{tan}\mathrm{4}\alpha{tan}\alpha=−\mathrm{7} \\ $$

Question Number 100575 Answers: 1 Comments: 0

Question Number 100570 Answers: 0 Comments: 1

Question Number 100567 Answers: 0 Comments: 3

{ ((x−(√(yz)) = 42)),((y−(√(xz)) = 6)),((z−(√(xy)) = −30)) :} find x+y+z =

$$\begin{cases}{{x}−\sqrt{{yz}}\:=\:\mathrm{42}}\\{{y}−\sqrt{{xz}}\:=\:\mathrm{6}}\\{{z}−\sqrt{{xy}}\:=\:−\mathrm{30}}\end{cases} \\ $$$${find}\:{x}+{y}+{z}\:= \\ $$

Question Number 100565 Answers: 1 Comments: 1

Question Number 100562 Answers: 0 Comments: 0

Question Number 100561 Answers: 1 Comments: 1

Question Number 100557 Answers: 2 Comments: 0

Ω=∫_0 ^∞ (e^(ax) /(e^(bx) +1))dx, b>a

$$\Omega=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{e}^{{ax}} }{{e}^{{bx}} +\mathrm{1}}{dx},\:{b}>{a} \\ $$

Question Number 100538 Answers: 0 Comments: 1

Pg 1144 Pg 1145 Pg 1146 Pg 1147 Pg 1148 Pg 1149 Pg 1150 Pg 1151 Pg 1152 Pg 1153

Contact: info@tinkutara.com