Sick-quences!! (6)

Algebra Level 4

T r T_ r is a sequence such that

∑ r = 1 n T r = n ( n + 1 ) ( n + 2 ) ( n + 3 ) 8 . \sum_{r =1 } ^ n T_r = \frac{ n(n+1)(n+2)(n+3) } { 8 }.

∑ r = 1 11 1 T r = a b \displaystyle \sum_{r=1}^{11} \frac{1}{T_r } = \frac{a}{b} , where a , b a, b are coprime positive integers. Find a + b a + b .


The answer is 233.

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Jaiveer Shekhawat by jaiveer shekhawat , 22, India

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4 solutions

Ayush Verma
Jan 19, 2015

T n = ∑ r = 1 n T r − ∑ r = 1 n − 1 T r = n ( n + 1 ) ( n + 2 ) ( n + 3 ) − ( n − 1 ) n ( n + 1 ) ( n + 2 ) 8 = n ( n + 1 ) ( n + 2 ) { ( n + 3 ) − ( n − 1 ) } 8 = n ( n + 1 ) ( n + 2 ) 2 ⇒ 1 T r = 2 r ( r + 1 ) ( r + 2 ) = 1 r − 2 r + 1 + 1 r + 2 = ( 1 r − 1 r + 1 ) + ( 1 r + 2 − 1 r + 1 ) ⇒ ∑ r = 1 11 1 T r = ( 1 1 − 1 12 ) + ( 1 13 − 1 2 ) = 77 156 ⇒ a + b = 77 + 156 = 233 { T }_{ n }=\sum _{ r=1 }^{ n }{ { T }_{ r } } -\sum _{ r=1 }^{ n-1 }{ { T }_{ r } } \\ \\ =\cfrac { n\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) -\left( n-1 \right) n\left( n+1 \right) \left( n+2 \right) }{ 8 } \\ \\ =\cfrac { n\left( n+1 \right) \left( n+2 \right) \left\{ \left( n+3 \right) -\left( n-1 \right) \right\} }{ 8 } \\ \\ =\cfrac { n\left( n+1 \right) \left( n+2 \right) }{ 2 } \\ \\ \Rightarrow \cfrac { 1 }{ { T }_{ r } } =\cfrac { 2 }{ r\left( r+1 \right) \left( r+2 \right) } \\ \\ =\cfrac { 1 }{ r } -\cfrac { 2 }{ r+1 } +\cfrac { 1 }{ r+2 } \\ \\ =\left( \cfrac { 1 }{ r } -\cfrac { 1 }{ r+1 } \right) +\left( \cfrac { 1 }{ r+2 } -\cfrac { 1 }{ r+1 } \right) \\ \\ \Rightarrow \sum _{ r=1 }^{ 11 }{ \cfrac { 1 }{ { T }_{ r } } } =\left( \cfrac { 1 }{ 1 } -\cfrac { 1 }{ 12 } \right) +\left( \cfrac { 1 }{ 13 } -\cfrac { 1 }{ 2 } \right) \\ \\ \quad =\cfrac { 77 }{ 156 } \\ \\ \Rightarrow a+b=77+156=233

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Jaiveer Shekhawat
Jan 17, 2015

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Ridiculously overrated problem. Should be around level 3 or 4

Rohit Shah - 6 years, 4 months ago

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For the same reason I re-checked it again and again if my answer is wrong because I couldn't believe this is a 280 point problem.

Kartik Sharma - 6 years, 4 months ago

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Well, next time I'm going to give you a problem of your level ... good things come later!!

jaiveer shekhawat - 6 years, 4 months ago

I've updated the problem to level 2.

Jake Lai - 6 years, 4 months ago

Very true.

Keshav Tiwari - 6 years, 4 months ago

Level 2 or 3 is ok.

Kaushal Agrawal - 6 years, 4 months ago

How did you get that striking with LATEX?

What Sorcery is that ? :O

Priyatam Roy - 6 years, 4 months ago

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Well, It's the magic of MS WORD. It's a lot simpler than latex.

jaiveer shekhawat - 6 years, 4 months ago

isn't the sum 121/273?

Sanjoy Roy - 6 years, 4 months ago
Abhijeet Verma
Feb 14, 2015

it would be easier to solve it by finding the general term first , that is Tr which would be n(n+1)(n+2)/2. then we can find the required sum for 1/Tr.

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Lu Chee Ket
Feb 1, 2015

S (n) = {3, 15, 45, 105, 210, 378, 630, 990, 1485, 2145, 3003...}

T (1) = S (1), T (2) = S (2) - S (1) and etc. Therefore,

T (n) = {3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858...}

1/ T (n) = {1/ 3, 1/ 12, 1/ 30, 1/ 60, 1/ 105, 1/ 168, 1/ 252, 1/ 360, 1/ 495, 1/ 660, 1/ 858...}

Sum of T (n) from n = 1 to 11 = 0.493589743589744+ = 77/ 156

77 + 156 = 233

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Who will go all the way to 15 DECIMAL PLACES !!

jaiveer shekhawat - 6 years, 4 months ago

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