If you are specially interested only in $\zeta(4)$, the following proof would work but this is an adaptation Euler''s idea. The idea is just to mimic Euler''s proof for the Basel problem. Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$

*A2A Now, [math](x^4+x^2+1)[/math][math]=(x^4+2x^2+1)-x^2=(x^2+1)^{2}-(x)^{2}[/math][math]=(x^2+x+1)(x^2-x+1)[/math] A keen observation reveals : [math](x^3+1)=(x+1

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It has long been established that ortho-disubstituted benzenes (for example, 1,2 and 1,6) exist in only one form, whereas Kekulé’s formula permits isomerism of such compounds (substituents at the carbon atoms linked with single or double bonds).

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Ln(1 + x) = x - x²/2 + x³/3 - (x^4)/4+……. Putting x = 1 The series becomes 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6……. So as x =1 Put it in ln(x +1) We get ln2 = 0.693

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Please see the explanation below The general equation of a hyperbola is (x-h)^2/a^2-(y-k)^2/b^2=1 Here, The equation is (x-1)^2/2^2-(y+2)^2/3^2=1 a=2 b=3 c=sqrt(a^2+b^2)=sqrt(4+9)=sqrt13 The center is C=(h,k)=(1,-2) The vertices are A=(h+a,k)=(3,-2) and A''=(h-a

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