Recursion formula for: $a_n = \left(\frac12\right)^n+2n−1$

Question

How do I write a recursive formula for: $$a_n = \left(\frac12\right)^n+2n−1$$ so we were taught to expand the recursion and the multiply all together. So is the answer:

$$a_1=\frac32,\quad a_n=a_{n-1}+\frac52$$

How do you show the expanding and multiplying?

Answer 1

You have $a_1=\dfrac{1}{2} + 2 - 1=\dfrac{3}{2}$

And for all $n \geq 0$, $a_{n+1}=\dfrac{1}{2}(n+1)+2(n+1)-1 = \dfrac{1}{2}n + \dfrac{1}{2} +2n+1 = \left(\dfrac{1}{2}n + 2n - 1 \right) + 2 + \dfrac{1}{2} = a_n + \dfrac{5}{2}$

Then for $n \geq 1$, $a_n=a_{n-1}+\dfrac{5}{2}$

Answer 2

This answer is probably not correct.

Going from explicit to recursive is easy for any sequence; just compute $a_{n+1} - a_n$:

$a_{n+1} - a_n = \left[\left(\dfrac {1}{2} \right)^{n+1}+2(n+1)-1\right] - \left[\left( \dfrac {1}{2} \right)^n+2n-1 \right]$

There we go, we've already god a recursive sequence. Can you simplify it from here?

EDIT: To simplify $a_n-a_{n-1} = \left(\dfrac12 \right)^n - \left( \dfrac12 \right)^{n-1}+ 2$

Rewrite $\left(\dfrac12 \right)^n$ as $\left(\dfrac12 \right)^{n-1} \cdot \left(\dfrac12 \right)^1$

$$\left(\dfrac12 \right)^{n-1} \cdot \left( \dfrac 12\right) - \left( \dfrac12 \right)^{n-1}+ 2$$

The first two terms both have $\left(\dfrac12 \right)^{n-1}$, so factor it out:

$$ \left( \dfrac {1}{2} \right)^{n-1} \left(\dfrac 12 - 1 \right) +2 $$

$\dfrac 12 - 1$ is equal to $- \dfrac 12$

$$ = \left( \dfrac {1}{2} \right)^{n-1} \left(-\dfrac 12 \right)^1 +2$$

Take the negative outside and add the powers

$$ -\left(\dfrac {1}{2} \right)^{n}+2$$

Answer 3

$$a_n = (\frac 12)^n + 2n − 1\\ a_{n+1} = (\frac 12)^{n+1} + 2n + 1\\ a_{n+1}=\frac12(\frac 12)^{n} + 2n + 1\\ a_{n+1}=\frac12((\frac 12)^{n} + 4n+2)=\\ a_{n+1}=\frac12((\underbrace{\frac 12)^{n} + 2n-1}_{a_n}+2n+3)=\\ a_{n+1}=\frac12(a_n+2n+3)=\\ $$ now you can be sure that $a_{n+1}$ contain a term such as $\frac12 a_n$ but you can get $$a_{n+1}=\frac12 a_n +an +b$$ then find a,b by initial conditions $$a_2=\frac{1}{4}+2(2)-1 \to a_2=\frac12 (a_1)+a(2)+b\\a_3=(\frac12)^3 +2(3)-1 \to a_3=\frac{1}{2}a_2+3a+b$$ find $a,b$ form system of equation ,finnaly you have the relation