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Mathematics 8 Online
OpenStudy (samigupta8):

If H1,H2,H3,....,H2n+1 are in H.P. Then summation of (-1)^i (Hi+Hi+1/Hi-Hi+1) from i =1 to 2n..is

OpenStudy (samigupta8):

Hi+1 is just the symbol for the next term of H.P after Hi

OpenStudy (samigupta8):

@freckles

OpenStudy (samigupta8):

@ganeshie8

OpenStudy (samigupta8):

@michele_laino

OpenStudy (samigupta8):

@hartnn

OpenStudy (samigupta8):

@parthkohli

Parth (parthkohli):

Let \(a_1, a_2, a_3, \cdots, a_{2n+1}\) be the corresponding reciprocals.\[\sum_{i=1}^{2n}(-1)^i\frac{\frac{1}{a_i}+\frac{1}{a_{i+1}}}{\frac{1}{a_i}-\frac{1}{a_{i+1}}}\]\[= \sum_{i=1}^{2n}(-1)^i \frac{ a_{i+1} + a_i}{a_{i+1}-a_{i}}\]\[= \frac{1}d \sum_{i=1}^{2n} (-1)^i (a_{i+1}+ a_i )\]\[= \frac{1}d \left(-a_{1}-a_{2} + a_2 + a_3 - a_3- a_4 + a_4 + a_5 + \cdots +a_{2n}+a_{2n+1} \right)\]\[= \frac{1}d (a_{2n+1}-a_1)\]\[= \frac{1}d (a_1 + 2nd - a_1)\]\[ = 2n \]

OpenStudy (samigupta8):

Wow!!! @parthkohli. .. Thanks a lot!!! :)

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