Think in trigo!

\mathfrak T=\cot \underbrace{\left [ \sum_{n=1}^{23} \text{arccot} \left( 1 + \displaystyle \mathop{\; \sum}_{k=1}^n 2k \right) \right ] }_{\color{#D61F06}{\mathfrak J}}

For $\color{#D61F06}{\mathfrak J}$ , use: $(1).~\displaystyle\sum_{r=1}^k r=\left(\dfrac{k(k+1)}{2}\right)\\ (2).~\cot^{-1}x=\tan^{-1}\left(\frac 1x\right)$ So that ,

$\color{#D61F06}{\mathfrak J}=\left(\sum_{n=1}^{23}\tan^{-1}\left(\dfrac{1}{1+n(n+1)}\right)\right)$

$=\left(\sum_{n=1}^{23}\tan^{-1}(n+1)-\tan^{-1}n\right)$

$\mathbf{A~Telescopic ~Series}$

$=\tan^{-1}(24)-\tan^{-1} 1=\tan^{-1}\left(\dfrac{23}{25}\right)=\cot^{-1}\left(\dfrac{25}{23}\right)$

So,

$\mathfrak T=\cot \color{#D61F06}{\mathfrak J}=\boxed{\dfrac{25}{23}}$

putting n=1 we get, cot [cot^(-1) (1+2)]

similarly, putting n as 3 we get cot [ cot^(-1) (3) + cot^(-1) (7) + cot^(-1) (13) ]

finally putting n =23, we get : cot [cot^(-1) (3) + cot^(-1) (7) +...............+cot^(-1) (553)]

we know that cot^(-1) (x) = tan^(-1) (1/x)

therefore above equation becomes

cot [ tan^(-1) (1/3) + .........(you got that)......+tan^(-1) (1/553)]

and we can write the following fractions as: 1/3 = (2-1)/(1+2 3) 1/7 = (3-2)/(1+3 2) 1/13 = (4-3)/(1+4 3) and so on till 1/553 = (24-23)/(1+24 23)

got the pattern!?!? wink

so tan^(-1) (b) - tan^(-1) (a) = tan^(-1) [(b-a)/(1+b*a)] implementing the above formula in the equation which we got earlier,.......(typing solutions is so tedious than solving it)

cot [ tan^(-1)(2) - tan^(-1) (1) +tan^(-1) (3) - tan^(-2) (2)+ tan^(-1) (4) - tan^(-1) (3)+ ..............+tan^(-1) (24) - tan^(-1) (23)]

so all other terms than ones containing 24 and 1get cancelled {I DON'T KNOW HOW TO DO THAT THINGY OF CANCELLATION} ok back to the problem

we get cot [tan^(-1) (24) - tan^(-1) (1)] which is equal to cot [ tan^(-1) (23/25)]

cot = 1/tan therefore cot [ tan^(-1) (23/25)] = 1/{tan[ tan^(-1) (23/25)]}

which gives us the answer 25/23.

STAY AWESOME