1. ## quick question

how could this one $(-3)(2)^{n-1} + 2^n(-1)$ simplify to $-2^{n+2}+3$?

2. Originally Posted by zpwnchen
how could this one $(-3)(2)^{n-1} + 2^n(-1)$ simplify to $-2^{n+2}+3$?
It doesn't. I'm getting
$-2^{n+2}+3.2^{n-1}$

3. Thanks! but i do not know why it was deduced to $-2^{n+2}+3$ in solution manual.

4. Originally Posted by zpwnchen
how could this one $(-3)(2)^{n-1} + 2^n(-1)$ simplify to $-2^{n+2}+3$?
It is not true for $n=2$. Is it?

5. it was consistently true for all n terms.

$a_n = 2a_{n-1} -3 , a_0=-1$
$(-3)(2)^{n-1} + 2^n(-1)$ --> $-2^{n+2}+3$

6. $\forall n \geq 2, \ -3 \cdot (2)^{n-1} + -2^n \equiv 0 (mod 2)$
But $-2^{n+2}+3 \equiv 1 (mod 2)$

So it is obviously not true.

7. Yeh! that's right!

I was just keep checking it " $-2^{n+2}+3$" against " $a_n = 2a_{n-1} -3 , a_0=-1$" for n terms and it was true.

So " $(-3)(2)^{n-1} + 2^n(-1)$" is not right! I suppose it must be something wrong in somewhere.

Thank you!

8. Originally Posted by zpwnchen
Yeh! that's right!

I was just keep checking it " $-2^{n+2}+3$" against " $a_n = 2a_{n-1} -3 , a_0=-1$" for n terms and it was true.

So " $(-3)(2)^{n-1} + 2^n(-1)$" is not right! I suppose it must be something wrong in somewhere.
For the sequence $a_n = 2a_{n-1} -3 , a_0=-1$ the function $f(n)=-2^{n+2}+3$ does give the terms of the sequence for $n=1,2,\cdots$.