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Algebra Level 3

S = 1 1 × 2 − 1 2 × 3 − 1 3 × 4 − ⋯ − 1 59 × 60 \large S = {\frac{1}{1 \times 2} - \frac{1}{2 \times 3} - \frac{1}{3 \times 4} - \cdots - \frac{1}{59 \times 60}}

Find 1 S \dfrac 1S .

Bonus: Generalise it for any n n , where n n is the last ending number, that is as 60 in the sum above.


The answer is 60.

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Kartik Jay by Kartik Jay , 19, India

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1 solution

Chew-Seong Cheong
Oct 30, 2017

S n = 1 1 × 2 − 1 2 × 3 − 1 3 × 4 − ⋯ − 1 ( n − 1 ) × n = 1 1 − 1 2 − 1 2 + 1 3 − 1 3 + 1 4 − ⋯ − 1 n − 1 + 1 n = 1 1 − 1 2 − 1 2 + 1 3 − 1 3 + 1 4 − 1 4 ⋯ 1 n − 1 − 1 n − 1 + 1 n = 1 n \begin{aligned} S_n & = {\color{#3D99F6}\frac 1{1\times 2}} - {\color{#D61F06}\frac 1{2\times 3}} - {\color{#3D99F6}\frac 1{3\times 4}} - \cdots - \color{#3D99F6} \frac 1{(n-1)\times n} \\ & = {\color{#3D99F6}\frac 11 - \frac 12} - {\color{#D61F06}\frac 12 + \frac 13} - {\color{#3D99F6}\frac 13 + \frac 14} - \cdots - \color{#3D99F6}\frac 1{n-1} + \frac 1n \\ & = \cancel{\frac 11 - \frac 12 - \frac 12} + \cancel{\frac 13 - \frac 13} + \cancel{\frac 14 -\frac 14} \cdots \cancel{\frac 1{n-1}- \frac 1{n-1}} + \frac 1n \\ & = \frac 1n \end{aligned}

⟹ 1 S n = n For n = 60 1 S 60 = 60 \begin{aligned}\implies \frac 1{S_n} & = n & \small \color{#3D99F6} \text{For }n = 60 \\ \frac 1{S_{60}} & = \boxed{60} \end{aligned}

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