# How can I find the general term of sequence with this relations?

It is given that $a_1=1$ and $a_2=0.5$ satisfying that for all integers $n \ge2$

$$n(n+1)a_{n+1}a_n+na_na_{n-1}=(n+1)^2a_{n+1}a_{n-1}$$

How can I found the general term of the sequence ${a_n}$ ?

I try to use the recurrence relations and think about using mathematical induction, but failed because of insufficient data.

Can I get some clues?

• What are the first few terms? – vadim123 Jul 25 '17 at 14:52
• @vadim123 $a_1=1$ and $a_2=0.5$ is given. – user362325 Jul 25 '17 at 14:52
• What are the next 10 terms? – vadim123 Jul 25 '17 at 14:53
• just curious, where is the problem from? – Tai Jul 25 '17 at 14:55
• user362325 I believe you asked this earlier. Did you delete the earlier question you wrote, in order to ask it again? In any case: you've earlier asked a question concluding "I try to use the recurrence relations and think about using mathematical induction, but failed because of insufficient data." – amWhy Jul 25 '17 at 15:07

Rearrange the equation, you get $$(n+1)^2\,a_{n+1}\,a_{n-1}-n(n+1)\,a_{n+1}\,a_n\,=n\,a_n\,a_{n-1}$$

$$(n+1)\,a_{n+1}\,\{(n+1)\,a_{n-1}-na_n\}=n\,a_n.a_{n-1}$$ $${(n+1)\,a_{n-1}\,-n\,a_n\over n.a_n.a_{n-1}}={1\over{(n+1)\,a_{n+1}}}$$ $${1\over a_n}-{1\over a_{n-1}}={1\over{(n+1)\,a_{n+1}}}\,\,-{1\over {n\,a_n}}$$

Now on putting the values $$n=2,3,......,n-1,n$$ And adding all the equations, you get

$${a_{n+1}\over a_n}={1\over(n+1)}$$ $${a_n\over{a_{n-1}}}\cdot{{a_{n-1}\over{a_{n-2}}}}\cdots\cdots {a_2\over a_1}={{1\over n}\cdot}{1\over{n-1}}\cdots\cdots{1\over2}\cdot{1\over1}$$ $$a_n={1\over n!}$$

• This solution is beautiful. – user362325 Jul 25 '17 at 16:50
• @user362325 It is unbecoming of you to "brag" or "boast" about an answer from a sockpuppet account of yours. – amWhy Jul 25 '17 at 17:26
• @amWhy: How can you be sure that there is a sockpuppet account of user362325? – user 170039 Jul 28 '17 at 3:12

Given that $a_1=1$, $a_2=\frac{1}{2}$. Putting $n=2,3,\dots~~$ we find that \begin{align*} a_3&=\frac{1}{6}=\frac{1}{3!}\\ a_4&=\frac{1}{2^3.3}=\frac{1}{4!}\\ a_5&=\frac{1}{2^3.3.5}=\frac{1}{5!}\\ a_6&=\frac{1}{2^4.3^2.5}=\frac{1}{6!} \end{align*} and so on. So, $$a_n=\frac{1}{n!}.$$ In fact, assuming $~a_n=\frac{1}{n!}~$ is true for some $n$, \begin{align*} (n+1)a_{n+1}\frac{1}{(n-1)!}+n\frac{1}{n!}\frac{1}{(n-1)!}&=(n+1)^2a_{n+1}\frac{1}{(n-1)!}\\ \implies\qquad (n+1)a_{n+1}+\frac{1}{((n-1)!}&=(n+1)^2a_{n+1}\\ \implies\qquad\qquad\qquad a_{n+1}&=\frac{1}{(n+1)!}. \end{align*}