How to see that $2^{n-1} + 2^{n-1} - 1 = 2^n - 1$

Question

How to see that $2^{n-1} + 2^{n-1} - 1 = 2^n - 1$?

Is there a rule about adding two powers of the same base I'm not aware of? I know that you can "add the exponents" if you are multiplying numbers of the same base, or "subtract" them if you are dividing.

$a+a=2\cdot a$ is true for all a; in this case, as the various answers have noted, you have $a=2^{n-1}$. — Steven Stadnicki, Mar 14 '15 at 16:54

score 3 · Accepted Answer · answered Mar 14 '15 at 16:53

3

The general rule is $$\underbrace{k^n+k^n+\cdots +k^n}_k=k\cdot k^n=k^1k^n=k^{1+n}$$

answered Mar 14 '15 at 16:53

vadim123

82,796

score 2 · Answer 2 · answered Mar 14 '15 at 16:52

2

$$2^{n-1} + 2^{n-1} - 1 = 2\cdot2^{n - 1}-1=2^{1+n-1}-1=2^n-1$$

answered Mar 14 '15 at 16:52

Adi Dani

16,949

score 1 · Answer 3 · answered Mar 14 '15 at 16:51

1

Simply $2^{n-1}+2^{n-1}=2\cdot 2^{n-1}=2^n$ which works for the base $2$ - for base three you'd need to add three times $3^{n-1}$.

answered Mar 14 '15 at 16:51

Mark Bennet

100,194

score 1 · Answer 4 · answered Mar 14 '15 at 16:51

1

$2^{n-1} + 2^{n-1} - 1 = 2\cdot2^{n-1} - 1 = 2^n - 1$

answered Mar 14 '15 at 16:51

Stefan Perko

12,467

How to see that $2^{n-1} + 2^{n-1} - 1 = 2^n - 1$

4 Answers4