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Question Number 219119 Answers: 4 Comments: 0

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Question Number 219113 Answers: 3 Comments: 1

((a + 3b)/(a + b−1)) + ((a + 3b−1)/(a + b−3)) = 4 ⇒ a + b = ?

$$\frac{\mathrm{a}\:+\:\mathrm{3b}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{1}}\:+\:\frac{\mathrm{a}\:+\:\mathrm{3b}−\mathrm{1}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{3}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$

Question Number 219112 Answers: 0 Comments: 0

Prove it: In triangle ABC, AB=c, BC=b, AC=a ab^2 c + abc^2 −a^2 bc ≥ tan (A/2) ((2S^3 )/((a+b)^2 −c^2 ))

$$\mathrm{Prove}\:\mathrm{it}: \\ $$$$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{AB}=\mathrm{c},\:\mathrm{BC}=\mathrm{b},\:\mathrm{AC}=\mathrm{a} \\ $$$$\mathrm{ab}^{\mathrm{2}} \mathrm{c}\:+\:\mathrm{abc}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{bc}\:\geqslant\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\:\frac{\mathrm{2S}^{\mathrm{3}} }{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} } \\ $$

Question Number 219110 Answers: 1 Comments: 0

find the nth term of x_(n+1) = x_n (2−x_n ) in x_1

$${find}\:{the}\:{nth}\:{term}\:{of}\:{x}_{{n}+\mathrm{1}} \:=\:{x}_{{n}} \left(\mathrm{2}−{x}_{{n}} \right) \\ $$$${in}\:{x}_{\mathrm{1}} \\ $$

Question Number 219098 Answers: 2 Comments: 0

ζ(α)=Σ_(n=1) ^(+∞) (1/n^α )

$$\zeta\left(\alpha\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\alpha} }\:\: \\ $$

Question Number 219093 Answers: 3 Comments: 1

Question Number 219090 Answers: 2 Comments: 0

Prove that the sequence a_n =(1/( ((n!))^(1/n) )) is decreasing.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:{a}_{{n}} =\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\:\mathrm{is}\:\mathrm{decreasing}. \\ $$

Question Number 219088 Answers: 0 Comments: 0

Question Number 219087 Answers: 1 Comments: 0

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Question Number 219085 Answers: 0 Comments: 0

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Question Number 219083 Answers: 0 Comments: 0

Question Number 219078 Answers: 1 Comments: 0

(1−(1/(2m))).Σ_(k=1) ^m (−1)^(k−1) ∙k∙(((m!)^2 )/((m−k)!(m+k)!))=(1/4) Proof this formula

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{m}}\right).\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \centerdot{k}\centerdot\frac{\left({m}!\right)^{\mathrm{2}} }{\left({m}−{k}\right)!\left({m}+{k}\right)!}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$

Question Number 219077 Answers: 0 Comments: 0

∫_0 ^(+∞) (((sin(n))/n))^m dn=π∙(m/2^m )∙Σ_(φ=0) ^(m/2) (−1)^∅ ∙(((n−2φ)^(m−1) )/((m−φ)!∙φ!)) Proof this formula

$$\int_{\mathrm{0}} ^{+\infty} \left(\frac{{sin}\left({n}\right)}{{n}}\right)^{{m}} {dn}=\pi\centerdot\frac{{m}}{\mathrm{2}^{{m}} }\centerdot\underset{\phi=\mathrm{0}} {\overset{{m}/\mathrm{2}} {\sum}}\left(−\mathrm{1}\right)^{\emptyset} \centerdot\frac{\left({n}−\mathrm{2}\phi\right)^{{m}−\mathrm{1}} }{\left({m}−\phi\right)!\centerdot\phi!}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Proof}\:{this}\:{formula} \\ $$

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