try this

wat is s₁ s₂

edited

u r right.how???

As far as I know , one way of directly solving these sum is to take help of FOURIER SERIES .

Expanding the fourier series f (x ) = 2 x - x² in ( 0 , 3 ) yields ,

2 x - x² = - \sum_{n = 1}^{}{\propto }\frac{9}{n^{2}\pi ^{2}}\; cos \frac{2n{\pi}x}{3}\; + \sum_{n=1}^{}{\propto } \frac{3}{n{\pi }}\; sin \frac{2n{\pi }x}{3}

Putting x = 3 / 2 , we get 3 - 94 = \frac{9}{\pi ^{2}} \times [- \sum_{n = 1}^{}{\propto }\frac{cos \;n{\pi }}{n^{2}} ]

Which in turn gives , 11² - 12² + 13² - .......... upto infinity = \frac{\pi ^{2}}{12}

Similarly , expanding f ( x ) = - π , if -π < x < 0

= x , if 0 < x < π

we get f ( x ) = -\frac{\pi }{4}\; - \;\frac{2}{\pi } \;[ cos x + \frac{cos \;3x}{3^{2}} + \frac {cos\;5x}{5^{2}} + ...... ] + 3\; sin \; x - \frac{sin \;2x}{2} + \frac{3\; sin \;3x}{3} - \frac{4\;sin\;4x}{4} + ......

Putting x = 0 ,we get f ( 0 ) = -Ï€4 - 2Ï€ [ 11² + 13² + 15² + ......]

Now f ( x ) is discontinuos at x = 0 , f (0^{-}) = \pi \;and\; f (0^{+})= 0

So \;\;f( 0 ) = \;\frac{1}{2}\; [{\; f(0^{-} ) + f(0^{+})\;]= \frac{1}{2}\;[ \pi \;+\;0 ] = \frac{\pi }{2}

Hence \frac{\pi }{2} = -\frac{\pi }{4} - \frac{2}{\pi }\;[\;\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}} + ......]

From where we get \;\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}}\;+...... = \frac{\pi ^{2}}{8}

So we get S₁ = \frac{\pi ^{2}}{8}

and S₂ = \frac{\pi ^{2}}{12}

So 486 x S₁ / S² = \frac{\pi ^{2}}{8}\times \frac{12}{\pi ^{2}}\times 486 = 729

I am not exactly sure abt this thing -- its too complicated -- so this post is more of a copy -

paste -- hope you don't mind