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Question: If H<sub>1</sub>, H<sub>2</sub>, H<sub>3</sub> ……..H<sub>2n+1</sub> are in H.P. then \(\sum _ { i =...

If H1, H2, H3 ……..H2n+1 are in H.P. then i=12n(1)i\sum _ { i = 1 } ^ { 2 n } ( - 1 ) ^ { i } (Hi+Hi+1HiHi+1)\left( \frac{H_{i} + H_{i + 1}}{H_{i} - H_{i + 1}} \right)is equal to –

A

2n –1

B

2n + 1

C

2n

D

2n + 2

Answer

2n

Explanation

Solution

Let 1Hi+1\frac{1}{H_{i + 1}}1Hi\frac{1}{H_{i}}= k

\ i=12n(1)i\sum _ { i = 1 } ^ { 2 n } ( - 1 ) ^ { i } (Hi+Hi+1HiHi+1)\left( \frac{H_{i} + H_{i + 1}}{H_{i} - H_{i + 1}} \right)

= (1Hi+1+1Hi)\left( \frac{1}{H_{i + 1}} + \frac{1}{H_{i}} \right)= 2n