If half the inverse of the golden ratio equals sin ( n π ) , what is n ?
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Are you a moderator or can you modify answers to have Latex script ?
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No, I'm not a moderator here. Anyone can type in Latex. If you want to learn the basics of Latex you can click here .
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I am trying to clean up the presentation of this problem Convergence of Triangular number sums. Let us mop it up !
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@Vijay Simha – I'll comment a proper write up here itself, you can copy-paste it there.
1 + 1/3^3 + 1/6^3 + 1/10^3 + 1/15^3 + 1/21^3 + 1/28^3 + 1/36^3 + 1/45^3 + .... = A
and that the sum of reciprocals of squares of all the triangular numbers converges to B
1 + 1/3^2 + 1/6^2 + 1/10^2 + 1/15^2 + 1/21^2 + 1/28^2 + 1/36^2 + 1/45^2 + ....= B
and A/B can be expressed as M*(P - pi^2)/(pi^2 - O) where M, O and P are all integers.
What is the value of M + O + P ?
If the sum of reciprocals of cubes of all the triangular numbers converges to the number A ,
A = 1 + 3 3 1 + 6 3 1 + 1 0 3 1 + 1 5 3 1 + 2 1 3 1 + 2 8 3 1 + 3 6 3 1 + 4 5 3 1 + ⋯
and the sum of reciprocals of squares of all the triangular numbers converges to the number B ,
B = 1 + 3 2 1 + 6 2 1 + 1 0 2 1 + 1 5 2 1 + 2 1 2 1 + 2 8 2 1 + 3 6 2 1 + 4 5 2 1 + ⋯
then B A can be expressed as M ( π 2 − O ) ( P − π 2 ) where M , O and P are positive integers. Find M + O + P .
If the sum of reciprocals of cubes of all the triangular numbers converges to the number *(A*),
*[A = 1 + \dfrac{1}{3^3} + \dfrac{1}{6^3} + \dfrac{1}{{10}^3} + \dfrac{1}{{15}^3} + \dfrac{1}{{21}^3} + \dfrac{1}{{28}^3} + \dfrac{1}{{36}^3} + \dfrac{1}{{45}^3} + \cdots *]
and the sum of reciprocals of squares of all the triangular numbers converges to the number *(B*),
*[B = 1 + \dfrac{1}{3^2} + \dfrac{1}{6^2} + \dfrac{1}{{10}^2} + \dfrac{1}{{15}^2} + \dfrac{1}{{21}^2} + \dfrac{1}{{28}^2} + \dfrac{1}{{36}^2} + \dfrac{1}{{45}^2} + \cdots *]
then *(\dfrac AB*) can be expressed as *(M \dfrac{\left( P - \pi^2 \right)}{\left( \pi^2 - O \right)}*) where *(M, O*) and *(P*) are positive integers. Find *(M+O+P*).
Note: Before copy-pasting the above Latex formatted write-up remove all the asterix symbols (*) and replace them with a backslash ( \ ).
Same as Mr. Tapas Mazumdar , but slight differently.
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The golden ratio is given by
ϕ = 2 1 + 5
Our problem wants us to find n such that
sin ( n π ) = 1 + 5 1 = 4 5 − 1 = 4 − 1 + 5
The result is standard and is equal to the value of sin 1 8 ∘ = sin ( 1 0 π ) . Thus n = 1 0 .