Question

prove mathematical induction (sum 2^0 + 2^1 + ...+ 2^n is 2^n+1 - 1 for all n>=1 how can i prove???

Answer 1

prove for case \(n=1\) is easy enough right?

Answer 2

now we assume it is true if \(n=k\) , that is, we assume
\[2^2+2^1+2^2+...+2^k=2^{k+1}-1\] and using that fact we want to show it is true for \(n=k+1\)

Answer 3

what does the \(k+1\) statement say?
it says \(2^0+2^1+2^2+...+2^k+2^{k+1}=2^{k+2}-1\)
so that is what we need to work towards

Answer 4

ok i understand it but how explain step wise ??

Answer 5

\[2^0+2^1+2^2+...+2^k+2^{k+1}=2^{k+2}-1\]
\[\overbrace{2^0+2^1+2^2+...+2^k}+2^{k+1}=2^{k+2}-1\]
by induction we can replace that part in indicated by the formula we get to assume

Answer 6

ok typo there sorry

Answer 7

u r great dude.....thanks

Answer 8

\[\overbrace{2^0+2^1+2^2+...+2^k}+2^{k+1}=\overbrace{2^{k+1}-1}+2^{k+1}\]

Answer 9

equating the stuff under the braces is the induction hypothesis
the rest is algebra or arithmetic. make the right hand side look like \(2^{k+2}-1\)

Answer 10

it always works this way with induction and summation formulas. just break off the last term to use induction on the rest up until the last term, then do some algebra

Answer 11

oh and yw, hope all steps are clear