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Original question is -

In the case below an initial term and a recursive formula are given. Find u( n )

$u( 1 ) = - 3, u( n ) = 2 u( n - 1 ) + 5$

I have tried applying my knowledge of Arithmetic Progression(AP), Quadratic Progression(QP), Geometric Progression(GP). And the most I could deduce from the sequence is that the difference bettween each term is multiplying by 2 hence it is a Geometric Sequence.

$-3, -1, 3, 11, 27, ...$

$-1 - (-3) = 2$

$3 - (-1) = 4$

$11 - 3 = 8$

$27 - 11 = 16$

The goal is to identify the general term of the sequence. I would love to see the steps too.

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    $\begingroup$ Please use MathJax. Here is a tutorial. $\endgroup$ Commented May 20 at 15:52

4 Answers 4

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Let us look at the differences of the terms. The first difference is $-1-(-3) = 2$, the second is $4$, the third is $8$, and so on. We can guess $$a_{n+1} - a_n = 2^n, \tag{1}$$ where $a_1 = -3$. Using $(1)$, we get $$a_{n+1} - a_1 = \sum_{j=1}^n 2^j = 2^{n+1} - 2,$$ so $$a_{n+1} = 2^{n+1} - 2 + a_1 = 2^{n+1}-5.$$ Replacing $n+1$ with $n$, we have $$a_{n} = 2^{n}-5,$$ for all $n\ge 1$.

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The title information leaves many other possibilities. For example, using interpolation we see that $$ u(x)=\frac{x^4 - 6x^3 + 23x^2 - 18x - 36}{12}. $$ Indeed, we have $u(1)=-3,u(2)=-1,u(3)=3,u(4)=11,u(5)=27$ and so on.

Of course, the recursion formula gives a contradiction here.

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Take $u_n=t_n-5$

$$t_n-5=2t_{n-1}-10+5\\ \ \\t_n=2t_{n-1}$$

Take $t_n=2^np_n$

$$2^np_n=2^np_{n-1}\implies p_n=p_{n-1}=\cdots=p_1$$

So back substituting $$u_n=2^np_1-5$$

We have $u_1=-3$ which gives $p_1=1$

$$u_n=2^n-5$$

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If the difference between each consecutive term in a sequence is in Geometric Progression, each term of the sequence is of the form $t_n = an + b + cr^{n-1}$ where a, b and c are constants.

r is the common ratio of the differences and n is the number of term

We have 3 unknowns a, b and c so we can form 3 equations using the given format

$ a + b + c $ = -3

$2a + b + 3c $ = -1

$3a + b + 9c $ = 1

Solving the equations , we get

a=0, b=-5 and c=2

Therefore, $t_n = 2(2^{n-1}) - 5$

which can be simplified as $t_n = 2^{n} -5$

This progression looks like a G.P. minus constant.

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  • $\begingroup$ if the difference is in GP then term is $ar^n+b$ if the second difference is in gp then it comes to $ar^n+bn+c$ its doesnt change much since you will get $b=0$ but just saying $\endgroup$
    – RandomGuy
    Commented May 21 at 1:16

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