Question

@sorour_n11

Answer 1

ga bura gozal giz

Answer 2

galdim

Answer 3

:)

Answer 4

galdin :)

Answer 5

\[S=\sum_{i=1}^{\infty} \frac{a_{i-1}}{a_{i}^2-a_{i-1}^2}\]duzde?

Answer 6

bali

Answer 7

xob degat ela\[\frac{a_{i-1}}{a_{i}^2-a_{i-1}^2}=\frac{a_{i-1}}{(a_{i}-a_{i-1})(a_{i}+a_{i-1})}=\frac{1}{2}(\frac{1}{a_{i}+a_{i-1}}-\frac{1}{a_{i}-a_{i-1}})\]

Answer 8

duzde

Answer 9

bura jan buldum

Answer 10

bidaki soalin ozunda yazip\[a_n=2a_{n-1}+a_{n-2}\]\[a_n-a_{n-1}=a_{n-1}+a_{n-2}\]banabarin baraye \(i\ge 2\) yazabulux\[a_i-a_{i-1}=a_{i-1}+a_{i-2}\]indi bunu goyoram osta gatiran rabetaya\[\frac{a_{i-1}}{a_{i}^2-a_{i-1}^2}=\frac{1}{2}(\frac{1}{a_{i}+a_{i-1}}-\frac{1}{a_{i-1}+a_{i-2}})\]

Answer 11

duzde

Answer 12

xob dar nahayat\[S=\sum_{i=1}^{\infty} \frac{a_{i-1}}{a_{i}^2-a_{i-1}^2}=\frac{a_{0}}{a_{1}^2-a_{0}^2}+\sum_{\color\red{i=2}}^{\infty} \frac{a_{i-1}}{a_{i}^2-a_{i-1}^2}\]va\[S=\frac{a_{0}}{a_{1}^2-a_{0}^2}+\frac{1}{2}\sum_{\color\red{i=2}}^{\infty} (\frac{1}{a_{i}+a_{i-1}}-\frac{1}{a_{i-1}+a_{i-2}})\]alan bidana soal serie telescope yadan galir?

Answer 13

yokh

Answer 14

masalan