Question

1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + ...2^n = 2^(n+1) -1. Prove

Answer 1

Geometric progression

Answer 2

http://fym.la.asu.edu/~tturner/MAT_117_online/SequenceAndSeries/geometric_sequence_17.gif

Answer 3

I was told to use induction. I used induction to n=k+1 and got stuck trying to solve.

Answer 4

you want to prove
\[\sum_{i=0}^{n} 2^n = 2^0 + 2^1 + 2^2 + .... + 2^n = 2^{n+1} -1\]
using induction

Answer 5

first make a base case
n = 0 so
\[2^0 = 2^{0+1} - 1 = 1 \] so its true then assume an arbitrary k where

\[\sum_{i=0}^{k}2^i = 2^{k+1} - 1\]
is true and prove for k+1

\[\sum_{i=0}^{k+1}2^i = 2^{(k+1)+1} - 1\]
do you understand what im doing so far?

Answer 6

yes

Answer 7

just want to make sure is the RHS really written as:
\[2^{n+1} - 1\]

Answer 8

oh nvm haha i figured it out

so you move on to the inductive step
first split the LHS

\[2^{k+1} + \sum_{i=0}^{k}2^i = 2^{k+2} - 1\]
then by inductive hypothesis:
\[2^{k+1} + (2^{k+1} - 1) = 2^{k+2} - 1\]
\[2^{k}2 + 2^{k}2 - 1 = 2^{k+2} - 1\]
\[2(2^{k} + 2^{k}) - 1 = 2^{k+2} - 1\]
\[2(2^{k+1}) - 1 = 2^{k+2} - 1\]
\[2^{k+2} - 1 = 2^{k+2} - 1\]
Since both sides are equal this holds for k+1 thus this holds for n+1 by induction