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The sequence is defined recursively. Find its general term. $t_1 = 3, t_n = 2(t_{n-1}) + 3n$

All I did was find the first couple of terms which are $3, 12, 33, 78, \dots$

The only thing common is that all are multiples of 3 but the sequence is neither arithmetic nor geometric. I've no idea how to solve a mixed sequence. Please help.

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  • $\begingroup$ See if this answer to a similar question helps. $\endgroup$ – Brian M. Scott May 19 '20 at 22:54
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    $\begingroup$ Do you know how to solve $t_n=4t_{n-1}-5t_{n-2}+2t_{n-3}$? $\endgroup$ – J. W. Tanner May 19 '20 at 23:39
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Subtract $t_{n-1}=2t_{n-2}+3n-3$ from $t_n=2t_{n-1}+3n,$ yielding $t_n-t_{n-1}=2t_{n-1}-2t_{n-2}+3$;

subtract $t_{n-1}-t_{n-2}=2t_{n-2}-2t{n-3}+3,$

yielding the homogenous recurrence $t_n-2t_{n-1}+t_{n-2}=2t_{n-1}-4t_{n-2}+2t_{n-3}$

or $t_n=4t_{n-1}-5t_{n-2}+2t_{n-3}.$

The characteristic equation is $x^3-4x^2+5x-2=(x-1)^2(x-2)$,

so the solution is $t_n=A2^n+Bn+C$,

with $2A+B+C=3$, $4A+2B+C=12$, and $8A+3B+C=33$;

i.e., $A=6, B=-3$, and $C=-6$; i.e., $t_n=6\times2^n-3n-6.$

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Let's just try it a bit to recognize a pattern: $$t_n=2t_{n-1}+3n=2(2t_{n-2}+3(n-1))+3n$$ $$=4t_{n-2}+(2+1)3n-3\cdot2$$ $$=4(2t_{n-3}+3(n-2))+(2+1)3n-3\cdot2=8t_{n-3}+(4+2+1)3n-3\cdot2-3\cdot4\cdot2$$ $$=8(2t_{n-4}+3(n-3))+(4+2+1)3n-3\cdot2-3\cdot4\cdot2$$ $$=16t_{n-4}+(8+4+2+1)3n-3\cdot2-3\cdot4\cdot2-3\cdot8\cdot3$$ So it looks like this: let $k<n$ be a positive integer, we have $$t_n=2^kt_{n-k}+(2^{k}-1)3n-3\sum_{i=1}^{k-1}\left(2^i\cdot i \right)$$ and let $$S=\sum_{i=1}^{k-1}\left(2^i\cdot i \right)=2\cdot1+4\cdot2+8\cdot3+\dots+2^{k-1}\cdot(k-1)$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2S=4\cdot1+8\cdot2+\dots+2^{k-1}(k-2)+2^k(k-1)$$ $$\implies S-2S=2+4+8+\dots+2^{k-1}-2^k(k-1)=2^k-2-2^k(k-1)$$ $$\iff S=2^k(k-2)+2$$ and it is easy to prove by induction that $$t_n=2^kt_{n-k}+(2^{k}-1)3n-3(2^k(k-2)+2)$$ for all integers $k$ where $0<k<n$.

Now substitute $k=n-1$ to get $$t_n=2^{n-1}t_1+(2^{n-1}-1)3n-3(2^{n-1}(n-3)+1)-3$$ $$t_1=3 \implies t_n=3(2^{n + 1}-(n+2))$$

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  • $\begingroup$ you must have made a mistake; your final formula for $t_n$ does not produce $3,12,33,78,...$ $\endgroup$ – J. W. Tanner May 20 '20 at 2:20
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    $\begingroup$ @J.W.Tanner I edited the formula right but I am trying to find the mistake, just give me some time $\endgroup$ – Anas A. Ibrahim May 20 '20 at 2:27
  • $\begingroup$ @J.W.Tanner I got where the mistake was and edited it, now the solution is undoubtedly complete except for the induction part. $\endgroup$ – Anas A. Ibrahim May 20 '20 at 2:34
  • $\begingroup$ now your answer agrees with mine ;-) $\endgroup$ – J. W. Tanner May 20 '20 at 2:40

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