Demonstration of sum of powers of $2$

Question

Theorem : For every natural number $p$: $$\sum^p_{i=0} 2^i = 2^{p+1}-1$$ I trieed to demonstrate the theorem using induction

Demonstration :

$1)$ If we have $p=0$ then we get $2^0=2^{0+1}-1$ that is always true. $2)$ Supposing that the first statement is true then we get that $$\sum^{p+1}_{i=0} 2^i = 2^{p+2}-1$$ Now we know that: $$\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1}$$ and if that is true, must be also true that: $$2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}$$ $$2^{p+2}-1-(2^{p+1}-1)=2^{p+2}-2^{p+1}=2^{p+1}(2^1-2^0)=2^{p+1}$$ QED

Is this a valid demonstration?

Answer 1

I think your first statement in part 2 puts the cart before the horse (you've assumed what you want to prove, or appear to).

The inductive step assumes that it holds for $p$:

$$\sum^p_{i=0} 2^i = 2^{p+1}-1$$

Now, given that, look at one side or the other for $p+1$. I'll look at the left side:

$$\sum^{p+1}_{i=0} 2^i = \left(\sum^p_{i=0} 2^i\right) + 2^{p+1}$$

This doesn't assume what is to be proved. I'm just splitting out the last term of the sum.

Now, substitute the assumed statement for $p$ on the right side:

$$\sum^{p+1}_{i=0} 2^i = 2^{p+1}-1 + 2^{p+1}$$

and simplify:

$$\sum^{p+1}_{i=0} 2^i = 2 \cdot 2^{p+1}-1 = 2^{p+2}-1 = 2^{(p+1)+1} - 1,$$

and you're done.

Answer 2

It looks like you are on the right track, but there are a few things to point out. This may be a matter of preference, but I think the induction step should be stated more clearly. Something like

Induction Step: Suppose $\sum^{p}_{i=0} 2^i=2^{p+1}-1$ holds for all $p \in \{1,2,\ldots, k\}$. Then...

And proceed from there. At this point, I'm not sure of the benefit of saying $$\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1} \implies 2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}$$

It appears to me that the above does not use the induction hypothesis, and any conclusion you reach seems a bit circular. What might be a better move is to write $$\sum^{k+1}_{i=0} 2^i= \sum^{k}_{i=0} 2^i+2^{k+1}$$ This is nice because you can say something about the quantity $\sum^{k}_{i=0} 2^i$ using the induction hypothesis. Can you take it from here?

Answer 3

Inductive hypothesis: $\quad\displaystyle\sum_{i=0}^p 2^i=2^{p+1}-1$ for some $p\ge 0$.

We must deduce that $\quad\displaystyle\sum_{i=0}^{p+1} 2^i=2^{p+2}-1$.

Decompose the sum: \begin{align*}\sum_{i=0}^{p+1} 2^i=2^{p+2}-1 &= \sum_{i=0}^p 2^i+2^{p+1}\\ &=2^{p+1}-1 +2^{p+1}\qquad\scriptstyle(\text{inductive hypothesis)}\\ &=2\cdot2^{p+1}-1=2^{p+2}-1. \end{align*}

Answer 4

Another way to look at it is $$\sum^{p+1}_{i=0} 2^i = 2 \sum^p_{i=0} 2^i +1$$ which can be used to go from $p$ to $p+1$.

Answer 5

One can prove $$ \sum_{i=1}^{p}2^i=2^{p+1}-1 $$ without induction. Actually the same proof works for any geometric progression $a, aq, aq^2,\ldots$ if $q\ne 1$. For instance, we want to calculate the sum of the first $p+1$ terms. Let $$ a+aq+aq^2+\cdots+ aq^{p}=X. \tag1 $$ Multiplying (1) with $q$ and adding $a$ on both sides we get $$a+q(a+aq+aq^2+\cdots+aq^p)=qX+a.$$ This gives $$a+aq+aq^2+\cdots+aq^p+aq^{p+1}=qX+a,$$ i.e., $$ X+aq^{p+1}=qX+a.$$ It follows that $$ a(q^{p+1}-1)=(q-1)X$$ and consequently $$X=a\frac{q^{p+1}-1}{q-1}. $$ In the above case we have $a=1$ and $q=2$: $$\sum_{i=1}^{p}2^i=X=2^{p+1}-1. $$