For every natural number n, 20 + 21 + ...+ 2n = 2n+1 − 1.

It's true for n = 1 to 5 as follows:

n=1: 20 + 21 = 3 = 22 − 1

n=2: 20 + 21 + 22 = 7 = 23 − 1

n=3: 20 + 21 + 22 + 23 = 15 = 24 − 1

n=4: 20 + 21 + 22 + 23 + 24 = 31 = 25 − 1

n=5: 20 + 21 + 22 + 23 + 24 + 25 = 63 = 26 − 1

The general formula for all n is proven by mathematical induction as follows and shown here.

Base
case: Set n = 0. Then 2

^{0}= 1 = 2^{1}− 1.
Induction
step: Show that if the equation holds for any particular n, which the
above
does,

it also holds for n+1.

Let n be any natural number and 2

it also holds for n+1.

Let n be any natural number and 2

^{0}+ 2^{1}+ ... + 2^{n}= 2^{n+1}− 1 is true.
Then
2

^{0}+ 2^{1}+ ...+ 2^{n}+ 2^{n+1}= ( 2^{0}+ 2^{1}+ ... + 2^{n}) + 2^{n+1}
=
( 2

^{n+1}- 1) + 2^{n+1}
=
2 * 2

=
2^{n+1}- 1^{n+2}- 1. QED

There is another proof as well. I will post it this weekend to give readers some time to try to find it on their own.

+++++++++++

March 16. The other proof follows. 2

Reverse the order of the sum and let x = 2:

20 + 21 + ... + 2n = (2n+1 − 1)÷(2 - 1) = 2n+1 − 1.

+++++++++++

March 16. The other proof follows. 2

^{0}+ 2^{1}+ ...+ 2^{n }is a sum of a geometric progression. Using long division, the quotient of (xn+1 − 1)÷(x - 1) is xn + xn-1 + xn-2 +... 1 = xn + xn-1 + xn-2 +...+ x0.Reverse the order of the sum and let x = 2:

20 + 21 + ... + 2n = (2n+1 − 1)÷(2 - 1) = 2n+1 − 1.

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