# Identify the sequence of $-3, -1, 3, 11, 27, ...$ [closed]

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.

• Please use MathJax. Here is a tutorial. Commented May 20 at 15:52

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 $$n\ge 1$$.

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.

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 $$p_1=1$$

$$u_n=2^n-5$$

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.

• 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 Commented May 21 at 1:16