Induction Proof for Sum of First n Powers of 2 (2^0 + 2^1 + ... + 2^n = 2^(n+1) - 1)

We prove the sum of powers of 2 is one less than the next powers of 2, in particular 2^0 + 2^1 + ... + 2^n = 2^(n+1) - 1. In the lesson I will refer to this as "the sum of the first n powers of 2", but note it is actually the first n+1 powers of 2 because 2^0 is included. Saying "the first n powers of 2" is just to explain we're going from 2^0 to 2^n in our sum. We'll prove this result using mathematical induction, and it's quite straightforward. We get some additional insight into the sum of powers of 2 using binary representations. #Proofs

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[sk_]

the inclusion of the binary example was brilliant on your part. thanks!

[thejacobwindsor]

can somebody find a way to explain this to me in a way that makes sense? i swear ive spent hours trying to figure this one topic out and i still dont get it and i have a test over this (and several other topics of the like) in a day and a half.

im not panicking. you're panicking.

[think_about_itt]

YOUR INDUCTION HOLDS TRUE FOR K+1 as INDUCTIVE HYPO

4:33 The sum = 2^(k) -1
Please note that the summision shall only take value upto 2^[k-1] not 2^(k).
Example:
For k=4

which dilutes to:


2^(0)+2^(1)+2^(2)+2^(3)=15
15=15

Note that here: last term of left hand side is 2^(k-1) {which is 2^3} not 2^k {as you have shown 2^k in the video}

I might be mistaken, I am a student. Please clarify onto that.
Thanks :)

[zeeshanrabbani]

Can we form a eqn for this series' sum