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Σ_(n=1) ^∞ (−1)^n ∙((2n−1)/((2n)!))∙((π/2))^(2n) =Σ_(n=1) ^∞ ((2n−1)/((2n)!))∙(−((π/2))^2 )^n =(2xD−1)∣_(x=π/2) Σ_(n=1) ^∞ (((−x^2 )^n )/((2n)!)) =(2xD−1)∣_(x=π/2) [Σ_(n=0) ^∞ (((−x^2 )^n )/((2n)!))−1] =(2xD−1)∣_(x=π/2) (cos x−1) =(−2xsin x−cos x+1)∣_(x=π/2) =1−π where is wrong?

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\frac{\mathrm{2n}−\mathrm{1}}{\left(\mathrm{2n}\right)!}\centerdot\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2n}} \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2n}−\mathrm{1}}{\left(\mathrm{2n}\right)!}\centerdot\left(−\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{n}} \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!} \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \left[\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}−\mathrm{1}\right] \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right) \\ $$$$=\left(−\mathrm{2xsin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \\ $$$$=\mathrm{1}−\pi \\ $$$$\mathrm{where}\:\mathrm{is}\:\mathrm{wrong}? \\ $$

Question Number 142627 Answers: 0 Comments: 0

Question Number 142624 Answers: 0 Comments: 0

fine the equation and the corresponding sketch of graph of the imageof the straight line joining (−1,−1) and (2,1) under the transformation equation w=(2+i)z

$${fine}\:{the}\:{equation}\:{and}\:{the}\:{corresponding} \\ $$$${sketch}\:{of}\:{graph}\:{of}\:{the}\:{imageof}\:{the} \\ $$$${straight}\:{line}\:{joining}\:\left(−\mathrm{1},−\mathrm{1}\right)\:{and} \\ $$$$\left(\mathrm{2},\mathrm{1}\right)\:{under}\:{the}\:{transformation}\:{equation} \\ $$$${w}=\left(\mathrm{2}+{i}\right){z} \\ $$

Question Number 142623 Answers: 2 Comments: 0

find the zero of z^3 +729=0 z∈C

$${find}\:{the}\:{zero}\:{of}\:{z}^{\mathrm{3}} +\mathrm{729}=\mathrm{0} \\ $$$${z}\in\mathbb{C} \\ $$

Question Number 142622 Answers: 1 Comments: 0

find z for which f(z)=(z/(sinz)) is undefined where z∈C

$${find}\:{z}\:{for}\:{which}\:{f}\left({z}\right)=\frac{{z}}{{sinz}}\:{is}\:{undefined} \\ $$$${where}\:{z}\in\mathbb{C} \\ $$

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Question Number 142595 Answers: 1 Comments: 0

Prove :: sec^2 x=4Σ_(n=0) ^∞ {(1/([(2n+1)π−2x]^2 ))+(1/([(2n+1)π+2x]^2 ))}

$$\mathrm{Prove}\:::\:\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}=\mathrm{4}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\left[\left(\mathrm{2n}+\mathrm{1}\right)\pi−\mathrm{2x}\right]^{\mathrm{2}} }+\frac{\mathrm{1}}{\left[\left(\mathrm{2n}+\mathrm{1}\right)\pi+\mathrm{2x}\right]^{\mathrm{2}} }\right\} \\ $$

Question Number 142618 Answers: 0 Comments: 2

In how many ways can committee of 5 be formed from a group of 11 people consisting of 4 teachers and 7 students if there is no restriction in the selection ? _______________________

$$\:\:\:{In}\:{how}\:{many}\:{ways}\:{can}\:{committee} \\ $$$${of}\:\mathrm{5}\:{be}\:{formed}\:{from}\:{a}\:{group}\: \\ $$$${of}\:\mathrm{11}\:{people}\:{consisting}\:{of}\:\mathrm{4}\:{teachers} \\ $$$${and}\:\mathrm{7}\:{students}\:{if}\:{there}\:{is}\:{no}\: \\ $$$${restriction}\:{in}\:{the}\:{selection}\:? \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

Question Number 142617 Answers: 0 Comments: 0

Question Number 142613 Answers: 0 Comments: 0

Evaluate ∫_0 ^1 ((log(x)log((x/(1−x))))/( (√(x/(1−x)))))dx

$$\boldsymbol{\mathrm{Evaluate}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{log}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right)}{\:\sqrt{\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}}}\boldsymbol{\mathrm{dx}} \\ $$$$ \\ $$

Question Number 142592 Answers: 1 Comments: 0

Is there any android apk compute generating function GF = Π_(i=1) ^(m) (Σ_(k=1) ^(n_i ) C_k ^( n_i ) x^k ) thank you so much

$${Is}\:{there}\:{any}\:{android}\:{apk} \\ $$$${compute}\:{generating}\:{function} \\ $$$${GF}\:=\:\underset{{i}=\mathrm{1}} {\overset{{m}} {\Pi}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}_{{i}} } {\Sigma}}{C}_{{k}} ^{\:{n}_{{i}} } \:{x}^{{k}} \right) \\ $$$${thank}\:{you}\:{so}\:{much} \\ $$

Question Number 142573 Answers: 2 Comments: 0

Question Number 142865 Answers: 1 Comments: 1

Let a_1 , a_2 , a_3 , ... be an arithmethic progression of positive real numbers. Then (1/( (√a_1 )+(√a_2 )))+(1/( (√a_2 )+(√a_3 )))+∙∙∙+(1/( (√a_(n−1) )+(√a_n )))= (A) ((n+1)/( (√a_1 )+(√a_n ))) (B) ((n−1)/( (√a_1 )+(√a_n ))) (C) (n/( (√a_1 )+(√a_n ))) (D) (n/( (√a_n )−(√a_1 )))

$$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:...\:\mathrm{be}\:\mathrm{an}\:\mathrm{arithmethic}\:\mathrm{progression}\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Then} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{{a}_{\mathrm{2}} }+\sqrt{{a}_{\mathrm{3}} }}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\:\sqrt{{a}_{{n}−\mathrm{1}} }+\sqrt{{a}_{{n}} }}= \\ $$$$\left(\mathrm{A}\right)\:\frac{{n}+\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{{n}−\mathrm{1}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }} \\ $$$$\left(\mathrm{C}\right)\:\frac{{n}}{\:\sqrt{{a}_{\mathrm{1}} }+\sqrt{{a}_{{n}} }}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{n}}{\:\sqrt{{a}_{{n}} }−\sqrt{{a}_{\mathrm{1}} }} \\ $$

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