Trig + summation = Easy

Geometry Level 4

3 n = 1 ( 1 π k = 1 cot 1 ( 1 + 2 r = 1 k r 3 ) ) n = ? \large3\sum_{n=1}^{\infty}\left({\dfrac{1}{\pi}\displaystyle\sum_{k=1}^{\infty}\cot^{-1}\left(1+2\sqrt{\sum_{r=1}^k r^3}\right)}\right)^n = \, ?


The answer is 1.

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2 solutions

Rishabh Jain
Apr 7, 2016

Oral problem if you proceed in the right direction.. :-)

T = 3 n = 1 ( 1 π k = 1 cot 1 ( 1 + 2 r = 1 k r 3 ) J ) n \mathfrak{T}=3\displaystyle\sum_{n=1}^{\infty}\left(\underbrace{\dfrac{1}{\pi}\displaystyle\sum_{k=1}^{\infty}\cot^{-1}\left(1+2\sqrt{\sum_{r=1}^k r^3}\right)}_{\large\color{#D61F06}{\mathfrak J}}\right)^n

For J \color{#D61F06}{\mathfrak J} use, ( 1 ) . r = 1 k r 3 = ( k ( k + 1 ) 2 ) 2 ( 2 ) . cot 1 x = tan 1 ( 1 x ) (1).~\displaystyle\sum_{r=1}^k r^3=\left(\dfrac{k(k+1)}{2}\right)^2\\ (2).~\cot^{-1}x=\tan^{-1}\left(\frac 1x\right) So that

J = 1 π ( k = 1 tan 1 ( 1 1 + k ( k + 1 ) ) ) \color{#D61F06}{\mathfrak J}=\dfrac{1}{\pi}\left(\sum_{k=1}^{\infty}\tan^{-1}\left(\dfrac{1}{1+k(k+1)}\right)\right)

= 1 π ( k = 1 tan 1 ( k + 1 ) tan 1 k ) =\dfrac{1}{\pi}\left(\sum_{k=1}^{\infty}\tan^{-1}(k+1)-\tan^{-1}k\right)

A T e l e s c o p i c S e r i e s \mathbf{A~Telescopic ~Series}

= 1 π ( π 2 π 4 ) = 1 4 =\dfrac{1}{\pi}\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)=\dfrac 14

T = 3 n = 1 ( 1 4 ) n \Large\therefore \mathfrak T=3\sum_{n=1}^{\infty}\left(\dfrac 14\right)^n

I n f i n i t e G P \mathbf{Infinite~GP}

= 3 ( 1 4 1 1 4 ) = 1 =3\left(\dfrac{\frac 14}{1-\frac 14}\right)=\huge \boxed{\color{#0C6AC7}{1}}

no doubt @Rishabh Cool it is a easy problem ,,, btw nice solution by you ,,

Rudraksh Sisodia - 5 years, 2 months ago

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Thanks!! :-)

Rishabh Jain - 5 years, 2 months ago
. .
Apr 30, 2021

The expression on the problem is too complex. (not complex number.)

3 n = 1 ( 1 π k = 1 cot 1 ( 1 + 2 r = 1 k r 3 ) ) n = 1 \displaystyle 3 \sum ^ { \infty } _ { n = 1 } \left ( \frac { 1 } { \pi } \sum ^ { \infty } _ { k = 1 } \cot ^ { -1 } \left ( 1 + 2 \sqrt { \sum ^ { k } _ { r = 1 } r ^ { 3 } } \right ) \right ) ^ { n } = 1 .

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