If $a_1 = 1$ and $a_{n + 1} = 2a_n + 5$ for $n = 1, 2,\ldots$, then what is $a_{100}$ equal to?

Question

If $a_1 = 1$ and $a_{n + 1} = 2a_n + 5$ for $n = 1, 2,\ldots$, then what is $a_{100}$ equal to?

The answer is $6 \times 2^{99} - 5$.

I need to know the method basically.

My approach:

$a_1=1$, $a_2=7$, $a_3=19$ and so on...

so the difference between the successive terms in the series $6$, $12$, $24$, $48$ is in GP

Now I need to know the method to proceed from here onward so that we can find the value of $a_{100}$

So please explain how to proceed further.

Answer 1

A simple way to solve such a recurrence is by not considering $a_n$ directly, but $b_n = a_n + c$ for a suitable constant $c$.

$$\begin{align} a_{n+1} &= 2 a_n + 5\\ \iff a_{n+1} + 5 &= 2 (a_n + 5), \end{align}$$

so with $b_n = a_n + 5$ we immediately see $b_n = 2^{n-1}\cdot b_1$, and that then translates to $a_n = 2^{n-1}\cdot (1+5) - 5$. (The given answer has the wrong sign for the $5$ summand, by the way.)

Generally, when you have a recurrence

$$a_{n+1} = c\cdot a_n + d$$

with $c \neq 1$, adding $\frac{d}{c-1}$ yields

$$a_{n+1} + \frac{d}{c-1} = c\cdot a_n + d + \frac{d}{c-1} = c\cdot a_n + \frac{c}{c-1}d = c\cdot\left(a_n + \frac{d}{c-1}\right).$$

Answer 2

$$ A(x) = \sum_{n=0}^\infty a_n x^n = a_0 + \sum_{n=1}^\infty a_n x^n $$

$$ = -2 + \sum_{n=1}^{\infty} (2a_{n-1} + 5 ) x^n = -2 + 2\sum_{n=1}^{\infty } a_{n-1} x^n + 5\sum_{n=1}^{\infty} x^n $$

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

$$ \therefore \ A(x) = \frac{7x-2}{(1-x)(1-2x)} $$

$$ = \frac{3}{1-2x} - \frac{5}{1-x} $$

$$ = \sum_{n=0}^{\infty} 3 \times 2^n \times x^n - \sum_{n=0}^{\infty} 5 \times x^n $$

$$ = \sum_{n=0}^{\infty} (3 \times 2^n - 5 ) x^n $$

$$ \therefore \ a_n = 3 \times 2^n -5 \Rightarrow a_{100} = 3 \times 2^{100} - 5 = 6 \times 2^{99} - 5 $$

Answer 3

$$a_{n+1}=2a_n+5,\quad a_{n+2}=2a_{n+1}+5,$$ hence $$a_{n+2}-a_{n+1}=2(a_{n+1}-a_{n})$$ or $$a_{n+2} = 3a_{n+1}-2a_n,$$ which homogenous. We seek roots of the characteristic polynomial $x^2-3x+2$, they are equal to $1$ and $2$, hence the solutions to the homogenous problem are $a_n=c_1\cdot 1^n+c_2\cdot 2^n=c_1 +c_2\cdot 2^n$.

Now we plug it into the original equation and obtain that

$$c_1 +c_2\cdot 2^{n+1} = 2 (c_1+c_2 2^n)+5,$$ which yields $c_1=-5$. Then again, we have initial conditions $-5+c_2\cdot 2^1=2 c_2-5 = 1,$ so the solutions are $a_n = 3\cdot 2^n-5$ and $a_{100}=6 \cdot 2^{99}-5$.