An algebra problem by Rahil Sehgal

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\times2} + \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}

The 50-term sums S S and T T are defined as above. The ratio S T \dfrac ST can be expressed as m n \dfrac mn , where m m and n n are coprime positive integers. Find m + n m+n .


The answer is 153.

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Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Relevant wiki: Telescoping Series - Sum

S = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ + 1 99 × 100 = ∑ k = 1 50 1 2 k ( 2 k − 1 ) = ∑ k = 1 50 ( 1 2 k − 1 − 1 2 k ) = ∑ k = 1 50 1 2 k − 1 − ∑ k = 1 50 1 2 k = 1 1 + 1 3 + 1 5 + . . . + 1 99 − 1 2 ∑ k = 1 50 1 k = ( 1 1 + 1 2 + 1 3 + . . . + 1 100 ) − ( 1 2 + 1 4 + 1 6 + . . . + 1 100 ) − 1 2 H 50 where H n is the n th harmonic number. = H 100 − 1 2 ( 1 1 + 1 2 + 1 3 + . . . + 1 50 ) − 1 2 H 50 = H 100 − H 50 \begin{aligned} S & = \frac 1{1\times2} + \frac 1{3\times 4} + \frac 1{5 \times 6} + \cdots + \frac 1{99 \times 100} \\ & = \sum_{k=1}^{50} \frac 1{2k(2k-1)} \\ & = \sum_{k=1}^{50} \left(\frac 1{2k-1} - \frac 1{2k} \right) \\ & = \sum_{k=1}^{50} \frac 1{2k-1} - \sum_{k=1}^{50} \frac 1{2k} \\ & = \frac 11 + \frac 13 + \frac 15 + ... + \frac 1{99} - \frac 12 \sum_{k=1}^{50} \frac 1k \\ & = \left(\frac 11 + \frac 12 + \frac 13 + ... + \frac 1{100}\right) - \left(\frac 12 + \frac 14 + \frac 16 + ... + \frac 1{100}\right) - \frac 12 \color{#3D99F6} H_{50} & \small \color{#3D99F6} \text{where }H_n \text{ is the }n^{\text{th}} \text{ harmonic number.} \\ & = H_{100} - \frac 12 \left(\frac 11 + \frac 12 + \frac 13 + ... + \frac 1{50}\right) - \frac 12 H_{50} \\ & = H_{100} - H_{50} \end{aligned}

T = 1 51 × 100 + 1 52 × 99 + ⋯ + 1 99 × 52 + 1 100 × 51 = 2 ( 1 51 × 100 + 1 52 × 99 + ⋯ + 1 74 × 77 + 1 75 × 76 ) = 2 ∑ k = 1 25 1 ( 50 + k ) ( 101 − k ) = 2 151 ∑ k = 1 25 ( 1 50 + k + 1 101 − k ) = 2 151 [ ( 1 51 + 1 52 + 1 53 + . . . + 1 75 ) + ( 1 100 + 1 99 + 1 98 + . . . + 1 76 ) ] = 2 151 ( H 100 − H 50 ) = 2 151 S \begin{aligned} T & = \frac 1{51 \times 100} + \frac 1{52 \times 99} + \cdots + \frac 1{99 \times 52} + \frac 1{100 \times 51} \\ & = 2 \left( \frac 1{51 \times 100} + \frac 1{52 \times 99} + \cdots + \frac 1{74 \times 77} + \frac 1{75 \times 76} \right) \\ & = 2 \sum_{k=1}^{25} \frac 1 {(50+k)(101-k)} \\ & = \frac 2{151} \sum_{k=1}^{25} \left( \frac 1 {50+k} + \frac 1{101-k} \right) \\ & = \frac 2{151} \left[ \left( \frac 1{51} + \frac 1{52} + \frac 1{53} + ... + \frac 1{75} \right) + \left( \frac 1{100} + \frac 1{99} + \frac 1{98} + ... + \frac 1{76} \right) \right] \\ & = \frac 2{151} (H_{100} - H_{50}) \\ & = \frac 2{151}S \end{aligned}

⟹ S T = 151 2 ⟹ m + n = 151 + 2 = 153 \implies \dfrac ST = \dfrac {151}2 \implies m + n = 151 + 2 = \boxed{153}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Salute your commitment to write such neat illustrative answers!

Siva Bathula - 4 years, 3 months ago

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