Finding the general term of a sequence that $a_1 = 2\,$ and $\,a_{n+1} = (2\cdot a_n) + 5$.

Question

In this problem, it is given that $a_{1} = 2\,$ and $\,(a_{n+1} = 2\cdot a_{n}+5)$. So the first terms of the sequence are $(2, 9, 23, 51, 107, 219,...)$. It is asked to find the general term $a_{n}$ of this sequence of natural numbers. (It is clear that it is not to use the previous term as already given)

Answer 1

$a_1 = 2\,$ and $\,a_{n+1} = 2a_n + 5$

The above recurrence relation is non-homogeneous.

The general solution/closed form is given by (1) ;

$$\boxed{a_n=a_n^H+a_n^p}\tag1$$

Solving the $a_n^H$ part;

Using the characteristic equation method;

$$x-2=0 \implies a_n^H=c_12^n$$

Solving the $a_n^p$ part;

Put $a_n=P_0$ and $a_{n+1}=P_0$ in the recurrence relation;

$$P_0=2P_0+5\implies a_n^p=-5$$

Substituting in (1);

$$a_n=c_12^n-5$$

Using $a_1=2$, we get $c_1=\frac{7}{2}$;

$$\boxed{a_n=(7)2^{n-1}-5}$$

$$OR$$

Use of generating functions is an alternate to getting to the closed form;

We start by assuming a function, say $Q(x)$ where,

$$Q(x)=\sum_{n=0}^{\infty} a_nx^n=a_0+a_1x+a_2x^2+\cdots$$

$$\,a_{n+1} = 2a_n + 5$$

Multiplying by $x^n$ and applying $\sum_{n=0}^{\infty}$ on the above relation;

$$\sum_{n=0}^{\infty} a_{n+1}x^n = 2\sum_{n=0}^{\infty}a_nx^n + 5\sum_{n=0}^{\infty}x^n$$

$$\frac{1}{x}(Q(x)-a_0) = 2Q(x)+ \frac{5}{1-x}$$

$$a_0=\frac{-3}{2}$$

$$Q(x)=\frac{-3}{2(1-2x)}+\frac{5x}{(1-x)(1-2x)}$$

Using Partial Fractions;

$$Q(x)=\frac{-3}{2(1-2x)}-\frac{5}{1-x}+\frac{5}{1-2x}$$

Putting in $$Q(x)=\sum_{n=0}^{\infty} a_nx^n$$

$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$$

$$\frac{1}{1-2x}=\sum_{n=0}^{\infty} 2^nx^n$$

$$\sum_{n=0}^{\infty} a_nx^n=\frac{-3}{2}\sum_{n=0}^{\infty} x^n-5\sum_{n=0}^{\infty} x^n+5\sum_{n=0}^{\infty} 2^nx^n$$

Removing out the sum and $x^n$, we get

$$\boxed{a_n=(7)2^{n-1}-5}$$

Answer 2

Hint

$$a_n=b_n+k \quad \implies b_{n+1}=2b_n+(k+5)$$ Do you see what happens if you select $k=-5$ ?

Solve it for $b_n$, go back to $a_n$ and use the initial condition.

Answer 3

$a_3 = 23 = 2(2(2) + 5) + 5 = 2^3 + 2 \cdot 5 + 5$

$a_4 = 2(2(2(2) + 5) + 5) + 5 = 2^4 + 2^2 \cdot 5 + 2 \cdot 5 + 2^0 \times 5$

$a_n = 2^n + 2^{n - 2} \cdot 5 + ... + 2^1 \cdot 5 + 2^0 \cdot 5$

Proviso: Exclude negative exponents

$a_1 = 2^1 = 2$