Where's my Telescope?

Algebra Level 4

S = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ + 1 99 × 100 T = 1 51 × 100 + 1 52 × 99 + 1 53 × 98 + ⋯ + 1 100 × 51 \begin{aligned} S & = \frac 1{1\times 2} +\frac 1{3\times 4} +\frac 1{5\times 6} + \cdots + \frac 1{99\times 100} \\ T & = \frac 1{51\times 100} + \frac 1{52\times 99 } + \frac 1{53\times 98} + \cdots + \frac 1{100\times 51} \end{aligned}

If S T = p q \dfrac ST = \dfrac pq , where p p and q q are coprime positive integers. Find p + q p+q .


The answer is 153.

Aditya Dhawan by Aditya Dhawan , 20, India

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2 solutions

Chew-Seong Cheong
Sep 19, 2016

S = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ + 1 99 × 100 = 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ + 1 99 − 1 100 = 1 1 + 1 2 + 1 3 + ⋯ + 1 100 − 2 ( 1 2 + 1 4 + 1 6 + ⋯ + 1 100 ) = 1 1 + 1 2 + 1 3 + ⋯ + 1 100 − ( 1 1 + 1 2 + 1 3 + ⋯ + 1 50 ) = 1 51 + 1 52 + 1 53 + ⋯ + 1 100 \begin{aligned} S & = \color{#3D99F6}{\frac 1{1\times 2}} + \color{#D61F06}{\frac 1{3\times 4}} + \color{#3D99F6}{\frac 1{5\times 6}} + \cdots + \color{#D61F06}{\frac 1{99\times 100}} \\ & = \color{#3D99F6}{\frac 11- \frac 12} + \color{#D61F06}{\frac 13- \frac 14} + \color{#3D99F6}{\frac 15 - \frac 16} + \cdots + \color{#D61F06}{\frac 1{99} - \frac 1{100}} \\ & = \frac 11 + \frac 12 + \frac 13 + \cdots + \frac 1{100} - 2\left(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac 1{100} \right) \\ & = \frac 11 + \frac 12 + \frac 13 + \cdots + \frac 1{100} - \left(\frac 11 + \frac 12 + \frac 13 + \cdots + \frac 1{50} \right) \\ & = \frac 1{51} + \frac 1{52} + \frac 1{53} + \cdots + \frac 1{100} \end{aligned}

T = 1 51 × 100 + 1 52 × 99 + 1 53 × 98 + ⋯ + 1 98 × 53 + 1 99 × 52 + 1 100 × 51 = 2 51 × 100 + 1 52 × 99 + 1 53 × 98 + ⋯ + 2 75 × 76 = 2 151 ( 1 51 + 1 100 + 1 52 + 1 99 + 1 53 + 1 98 + ⋯ + 1 75 + 1 76 ) = 2 151 ( 1 51 + 1 52 + 1 53 + ⋯ + 1 100 ) = 2 151 S \begin{aligned} T & = \frac 1{51 \times 100} + \color{#3D99F6}{\frac 1{52 \times 99}} + \color{#D61F06}{\frac 1{53 \times 98}} + \cdots + \color{#D61F06}{\frac 1{98 \times 53}} + \color{#3D99F6}{\frac 1{99 \times 52}} + \frac 1{100 \times 51} \\ & = \frac 2{51 \times 100} + \color{#3D99F6}{\frac 1{52 \times 99}} + \color{#D61F06}{\frac 1{53 \times 98}} + \cdots + \frac 2{75 \times 76} \\ & = \frac 2{151}\left(\frac 1{51} + \frac 1{100} + \color{#3D99F6}{\frac 1{52} + \frac 1{99}} + \color{#D61F06}{\frac 1{53} + \frac 1{98}} + \cdots + \frac 1{75} + \frac 1{76} \right) \\ & = \frac 2{151}\left(\frac 1{51} + \frac 1{52} + \frac 1{53} + \cdots + \frac 1{100} \right) \\ & = \frac 2{151} S \end{aligned}

⟹ S T = 151 2 ⟹ p + q = 151 + 2 = 153 \implies \dfrac ST = \dfrac {151}2 \implies p+q = 151 + 2 = \boxed{153}

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Aditya Dhawan
Sep 19, 2016

S = ∑ k = 1 50 1 ( 2 k ) ( 2 k − 1 ) = ∑ k = 1 50 1 2 k − 1 − 1 2 k = ∑ k = 1 100 1 k − ∑ k = 1 50 1 k T = ∑ k = 1 50 1 ( 50 + k ) ( 101 − k ) = 1 151 ∑ k = 1 50 1 50 + k + 1 101 − k = 2 151 ∑ k = 1 50 1 50 + k = 2 151 { ∑ k = 1 100 1 k − ∑ k = 1 50 1 k } ∴ S T = 151 2 S=\quad \sum _{ k=1 }^{ 50 }{ \frac { 1 }{ (2k)(2k-1) } } =\quad \sum _{ k=1 }^{ 50 }{ \frac { 1 }{ 2k-1 } } -\quad \frac { 1 }{ 2k } =\quad \sum _{ k=1 }^{ 100 }{ \frac { 1 }{ k } } -\sum _{ k=1 }^{ 50 }{ \frac { 1 }{ k } } \\ \\ T=\quad \sum _{ k=1 }^{ 50 }{ \frac { 1 }{ (50+k)(101-k) } } =\frac { 1 }{ 151 } \sum _{ k=1 }^{ 50 }{ \frac { 1 }{ 50+k } } +\quad \frac { 1 }{ 101-k } \\ =\quad \frac { 2 }{ 151 } \sum _{ k=1 }^{ 50 }{ \frac { 1 }{ 50+k } } \\ =\frac { 2 }{ 151 } \left\{ { \sum _{ k=1 }^{ 100 }{ \frac { 1 }{ k } -\sum _{ k=1 }^{ 50 }{ \frac { 1 }{ k } } } } \right\} \\ \therefore \quad \frac { S }{ T } =\quad \frac { 151 }{ 2 }

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Your second terms for T T , 1 52 × 101 \dfrac 1{52 \times 101} is wrong. I will edit for you. You should also use three starting terms.

Chew-Seong Cheong - 4 years, 9 months ago

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Indeed. I have made the corrections. Thank you

Aditya Dhawan - 4 years, 9 months ago

Its not a 5 level question.

Priyanshu Mishra - 4 years, 8 months ago

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I have not marked the level.

Aditya Dhawan - 4 years, 8 months ago

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I am not saying to you. I have just said it without any complain.

Priyanshu Mishra - 4 years, 8 months ago

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