Answer by Boris Novikov for Recurrence Relation with Square Root
Hint:Denote $b_n=\sqrt{1+24a_{n}}$. Then you get $$4b_n^2-4=b_{n+1}^2+6b_{n+1}+5,$$whence$$2b_n=\pm (b_{n+1}+3).$$
View ArticleRecurrence Relation with Square Root
Well, I was doing a problem on recurrence relation , where there was given a an recurrence relation and we had to find $a_{n}$ or simplify the recurrence.The recurrence relation was$$\begin{align}...
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