$a_{n}^{2} + a_{n} = 4S_{n} + 3 \ （n \geqslant 1）$，find $a_{n}$

Asked Feb 27 '20 at 09:41

Active Feb 27 '20 at 10:40

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Set sequence $a_{n} > 0(n \geqslant 1)$.$S_{n}$ is the sum of the preceding $n$ terms of the sequence $a_{n}$ and satisfies $a_{n}^{2} + a_{n} = 4S_{n} + 3$.

1)Find the general term formula of sequence $a_{n}$;

2)$b_{n} \overset{\text{def}}{=} \frac{1}{a_{n}a_{n+1}}$. Find the sum of the preceding $n$ entries of the sequence $b_{n}$.

Through the conditions, I could get the following formulas

\begin{align} a_{n+1}^{2} + a_{n+1} = 4S_{n+1} + 3 \ \ \ \ (1)\\ a_{n}^{2} + a_{n} = 4S_{n} + 3 \ \ \ \ (2) \end{align}

By $(1) - (2)$,I have got $a_{n+1}^2 - 3a_{n+1} = a_{n}^2 + a_{n}$,but I couldn't continue.

edited Feb 27 '20 at 10:40

Ivan Neretin

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asked Feb 27 '20 at 09:41

Nothing of L_j_j

1

Just for clarity, is $S_n=\sum\limits_{i=1}^n a_i$, or $S_n=\sum\limits_{i=1}^{n-1} a_i$? – Keen-ameteur Feb 27 '20 at 10:10
Since $S_1=a_1$, one can find that $a_1=\frac{1}{2} \left(3+\sqrt{21}\right)$, and then that $a_2 = \frac{1}{2} \left(3+\sqrt{45+8 \sqrt{21}}\right)$ ... are you sure the equation is correct? – Greg Martin Feb 27 '20 at 10:46
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Why don't you answer to Keen-ameteur ? Your question is indeed ambiguous. – Jean Marie Feb 27 '20 at 10:56
I have shown that $n \geqslant 1$,so $S_{n}$ should be $\sum_{i=1}^{n}a_{i}$.@Keen-ameteur – Nothing of L_j_j Feb 27 '20 at 11:34
I'm sure I copied the question right.But I couldn't prove that there was not a general term formula of this sequence ,either.@GregMartin – Nothing of L_j_j Feb 27 '20 at 11:43
This is a non-linear recurrence, and not of any of the solvable forms (as far as I know). – vonbrand Feb 28 '20 at 14:15

$a_{n}^{2} + a_{n} = 4S_{n} + 3 \ （n \geqslant 1）$，find $a_{n}$

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