A series of triplets

Calculus Level 5

S = 1 1 ⋅ 2 ⋅ 3 + 1 4 ⋅ 5 ⋅ 6 + 1 7 ⋅ 8 ⋅ 9 + 1 10 ⋅ 11 ⋅ 12 + 1 13 ⋅ 14 ⋅ 15 + ⋯ S=\dfrac { 1 }{ 1\cdot 2\cdot 3 } +\dfrac { 1 }{ 4\cdot 5\cdot 6 } +\dfrac { 1 }{ 7\cdot 8\cdot 9 } +\dfrac { 1 }{ 10\cdot 11\cdot 12 } +\dfrac { 1 }{ 13\cdot 14\cdot 15 } + \cdots

If S S can be expressed in the form π A − A ln ⁡ A 4 A \dfrac { \pi \sqrt { A } -A\ln { A } }{ 4A } with A A being a positive, prime integer, find A A .

This problem is original.


The answer is 3.

View wiki
Jonas Katona by Jonas Katona , 22, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hassan Abdulla
Jul 15, 2018

S = ∑ k = 0 ∞ 1 ( 3 k + 1 ) ( 3 k + 2 ) ( 3 k + 3 ) = 1 2 ∑ k = 0 ∞ 1 3 k + 1 − 2 3 k + 2 + 1 3 k + 3 S = 1 2 ∑ k = 0 ∞ ∫ 0 1 ( x 3 k − 2 x 3 k + 1 + x 3 k + 2 ) d x = 1 2 ∫ 0 1 ( ( 1 − x ) 2 ∑ k = 0 ∞ x 3 k ) d x S = 1 2 ∫ 0 1 1 − x x 2 + x + 1 d x = 1 2 ( − 1 2 ∫ 0 1 2 x + 1 − 3 x 2 + x + 1 d x ) = − 1 4 ( ∫ 0 1 2 x + 1 x 2 + x + 1 d x − 3 ∫ 0 1 1 x 2 + x + 1 d x ) S = − 1 4 ( ln ⁡ ( x 2 + x + 1 ) − 6 3 tan ⁡ − 1 ( 2 x + 1 3 ) ∣ 0 1 ) = π 3 − 3 ln ⁡ ( 3 ) 4 ⋅ 3 A = 3 S=\sum_{k=0}^\infty \dfrac { 1 }{(3k+1)(3k+2)(3k+3)}=\dfrac{ 1 }{2}\sum_{k=0}^\infty \dfrac{ 1 }{3k+1}-\dfrac{ 2 }{3k+2}+\dfrac{ 1 }{3k+3}\\ S=\dfrac{ 1 }{2}\sum_{k=0}^\infty\int_{0}^{1}\left ({x^{3k}-2x^{3k+1}+x^{3k+2}} \right )dx=\dfrac{ 1 }{2}\int_{0}^{1} \left ( \left ( 1-x \right )^2 \sum_{k=0}^\infty x^{3k} \right )dx \\ S=\dfrac{ 1 }{2}\int_{0}^{1} \dfrac{ 1-x }{x^2+x+1}dx=\dfrac{ 1 }{2}\left (\dfrac{ -1 }{2}\int_{0}^{1} \dfrac{ 2x+1-3 }{x^2+x+1}dx \right )=\dfrac{ -1 }{4}\left (\int_{0}^{1} \dfrac{ 2x+1 }{x^2+x+1}dx-3\int_{0}^{1} \dfrac{ 1 }{x^2+x+1}dx \right )\\ S=\dfrac{ -1 }{4} \left(\left . \ln\left ( x^2+x+1 \right )-\dfrac{ 6 }{\sqrt{3}}\tan^{-1}\left ( \dfrac{ 2x+1 }{\sqrt{3}} \right ) \right |_0^1 \right)=\dfrac{ \pi \sqrt{3} -3\ln(3)}{4\cdot 3} \\ A= 3

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...