Harmonically Complex!

Algebra Level 3

i = 1 200 ( 1 ) i H i + H i + 1 H i H i + 1 \large \sum_{i=1}^{200} (-1)^{i} \frac{H_{i} + H_{i+1}}{H_{i} - H_{i+1}}

H 1 , H 2 , H 3 , H 201 H_1, H_{2}, H_{3}, \cdots H_{201} above are in a harmonic progression . Find the value of the sum above.


The answer is 200.

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1 solution

S = i = 1 200 ( 1 ) i H i + H i + 1 H i H i + 1 Let H i = 1 a + ( i 1 ) d , where a = first term and d = common difference. = i = 1 200 ( 1 ) i 1 a + ( i 1 ) d + 1 a + i d 1 a + ( i 1 ) d 1 a + i d = i = 1 200 ( 1 ) i 2 a + 2 i d d d = i = 1 200 ( 1 ) i ( 2 a d + 2 i 1 ) = ( 2 a d 1 ) i = 1 200 ( 1 ) i + 2 i = 1 200 ( 1 ) i i = ( 2 a d 1 ) ( 0 ) + 2 i = 1 200 ( 1 ) i i = 2 ( 1 + 2 3 + 4 5 + 6 199 + 200 ) = 2 ( 1 + 1 + 1 + + 1 ) 100 × 1 = 200 \begin{aligned} S & = \sum_{i=1}^{200} (-1)^i \frac {H_i + H_{i+1}}{H_i - H_{i+1}} \quad \quad \small \color{#3D99F6} \text{Let } H_i = \frac 1{a+(i-1)d} \text{, where } a = \text{ first term and } d= \text{ common difference.} \\ & = \sum_{i=1}^{200} (-1)^i \frac {\frac 1{a+(i-1)d} + \frac 1{a+id}}{\frac 1{a+(i-1)d} - \frac 1{a+id}} \\ & = \sum_{i=1}^{200} (-1)^i \frac {2a+2id-d}d \\ & = \sum_{i=1}^{200} (-1)^i \left(\frac {2a}d + 2i -1 \right) \\ & = \left(\frac {2a}d -1 \right) {\color{#3D99F6} \sum_{i=1}^{200} (-1)^i} + 2 \sum_{i=1}^{200} (-1)^i i \\ & = \left(\frac {2a}d -1 \right) {\color{#3D99F6} (0)} + 2 \sum_{i=1}^{200} (-1)^i i \\ & = 2 ({\color{#3D99F6}-1+2}{\color{#D61F06}-3+4}{\color{#3D99F6}-5+6}-\cdots{\color{#D61F06}-199+200}) \\ & = 2 \underbrace{({\color{#3D99F6}1} + {\color{#D61F06}1}+{\color{#3D99F6}1}+\cdots+{\color{#D61F06}1})}_{100 \times 1} \\ & = \boxed{200} \end{aligned}

Lovely Solution

Vatsalya Tandon - 4 years, 6 months ago

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