Please explain this method to solve series? Note:- I can do this question with another method but i want to know about this method. But I am not able to understand from E.Q-(1) so please explain this method & there is no need to solve question . PLEASE EXPLAIN. Thanks in advance!!

Aug 8, 2018

Explanation:

As per eqn.(1), $\left(2 r - 3\right) {t}_{r} + 8 {t}_{r} = \left(2 r - 1\right) {t}_{r + 1}$.

$\therefore 8 {t}_{r} = \left(2 r - 1\right) {t}_{r + 1} - \left(2 r - 3\right) {t}_{r}$

Now, we want to express ${t}_{r}$ as $f \left(r\right) - f \left(r \pm 1\right)$.

So, let us see if we succeed if we take $f \left(r\right) = \left(2 r - 3\right) {t}_{r}$.

$f \left(r\right) = \left(2 r - 3\right) {t}_{r} \Rightarrow f \left(r + 1\right) = \left\{2 \left(r + 1\right) - 3\right\} {t}_{r + 1}$,

$i . e . , f \left(r\right) = \left(2 r - 3\right) {t}_{r} \Rightarrow f \left(r + 1\right) = \left(2 r - 1\right) {t}_{r + 1}$.

Thus, we have been successful to express ${t}_{r}$ as,

$f \left(r + 1\right) - f \left(r\right) , \text{ with, } f \left(r\right) = \left(2 r - 3\right) {t}_{r}$.