Question

solution of this question, step by step​

Answer 1

Answer:

$\implies (n-2)+\dfrac{1}{n-2}$

Step-by-step explanation:

Given,

First term = a$_{1}$ = 0

Let the common difference between the consecutive terms of this AP be d.

Therefore, using the properties of AP,

Second term = a$_{1}$ + d = 0 + d = d

Similarly,

Third term = 2d , fourth term = 3d and so on, thus, nth term = ( n - 1 )d & ( n - 1 )th term = ( n - 2 )d.

$\implies \bigg( \dfrac{ a _{3} }{ a _{2}} + \dfrac{a _{4}}{ a _{3}} + ... + \dfrac{a _{n}}{a _{n - 1}} \bigg) - a _{2} \bigg( \dfrac{1}{a _{2}} + \dfrac{1}{a _{3}} + ... + \dfrac{1}{ a _{n - 2}} \bigg) \\ \\ \\ \implies \bigg\{\dfrac{ 2d }{d} + \dfrac{3d}{2d} + ... + \dfrac{(n - 1)d}{(n - 2)d} \bigg\} - d \bigg\{\dfrac{1}{d} + \dfrac{1}{2d} + ... + \dfrac{1}{(n - 3)d} \bigg \} \\ \\ \\ \implies \bigg \{2 + \dfrac{3}{2} + \dfrac{4}{3} + ... + \dfrac{n - 1}{n - 2} \bigg \} - \bigg \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n - 3} \bigg \} \\ \\ \\ \implies \bigg \{(1 + 1) + \dfrac{2 + 1}{2} + \dfrac{3 + 1}{3} + ... + \dfrac{n - 2 + 1}{n - 2} \bigg \} - \bigg \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n - 3} \bigg \} \\ \\ \\ \implies \bigg \{(1 + 1) + \overline{1 + \dfrac{1}{2}} + \overline{1 + \dfrac{1}{3}} + ... + \overline{1 + \dfrac{1}{n - 2} } \bigg \} - \bigg \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + ... \dfrac{1}{n - 3} \bigg \} \\ \\ \\ \implies (1 + 1 + ... \: \bold{upto \: (n - 2)\: terms}) \: + \bigg \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + ... \dfrac{1}{n - 3} \bigg \} + \dfrac{1}{n-2}- \bigg \{ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + ... \dfrac{1}{n - 3} \bigg \} \\ \\ \\ \implies \dfrac{n - 2}{2} \bigg \{1+1\bigg \} + \dfrac{1}{n - 2} \\ \\ \\ \implies \dfrac{(n - 2)2}{2} +\dfrac{1}{n - 2}$

$\implies (n-2)+\dfrac{1}{n-2}$

Answer 2

Hi

hope it helps you......