Question

How do you prove this by induction...? Like step two after you sub in k+1 ?
(1+x)^2 is greater than or equal to 1+nx where x>-1

Answer 1

must be a typo here
is there some other \(n\)?

Answer 2

whoops sorry the ^2 is a ^n

Answer 3

for example, if \(n=100\) and \(x=2\) then there is no way that \[(1+2)^2\geq 1+100\times2\]

Answer 4

(1+x) ^n greater or equal to 1+ nk where x>-1

Answer 5

kk that makes more sense

Answer 6

you know how to do a proof by induction? this should not be hard if you know it

Answer 7

first you have to show it is true if \(n=1\) i.e. you have to show that
\[(1+x)^1\geq 1+1\times x\] which is pretty darn obvious because they are the same
ok so far?

Answer 8

?

Answer 9

uh i got up until k+1

Answer 10

ok hold on
first you get to ASSUME it is true if \(n=k\) i.e. you get to assume that
\[(1+x)^k\geq 1+kx\]

Answer 11

like after n=1
Assume true for some n=k after proving this
so consider (1+x)^k + (1+x) ^k+1 is greater than or equal to 1+kx + (1+x)^k+1 where x>-1

Answer 12

then using the fact that
\[(1+x)^k\geq 1+kx\] prove that
\[(1+x)^{k+1}\geq 1+(k+1)x\]

Answer 13

it is not plus, it is times

Answer 14

or was that the hint?

Answer 15

really? cause for another problem my teacher showed
original ...k^2 = (k(k+1) (2k+1))/6
then k^2 + (k+1) ^2 =( (k(k+1) (2k+1))/6 ) + (k+1)^2

Answer 16

on that problem, i am assuming that the left hand side was a SUM right?

Answer 17

\[\sum_{i=1}^k i^2=\frac{k(k+1)(2k+1)}{6}\]

Answer 18

yea

Answer 19

that is way you got to add the last term
you get to assume that
\[\sum_{i=1}^k i^2=\frac{k(k+1)(2k+1)}{6}\] then note that
\[\sum_{i=1}^{k+1} i^2=\sum_{i=1}^k i^2+(k+1)^2\]

Answer 20

you have assumed that \[\sum_{i=1}^k i^2=\color{red}{\frac{k(k+1)(2k+1)}{6}}\] which makes \[\sum_{i=1}^{k+1} i^2=\sum_{i=1}^k i^2+(k+1)^2=\color{red}{\frac{k(k+1)(2k+1)}{6}}+(k+1)^2\]

Answer 21

mhm

Answer 22

ok in this case two things are different
a) it is not a sum, it is a produce and
b) it is not an equality, it is an inequality

Answer 23

so we start by assuming that we know
\[(1+x)^k\geq 1+kx\]

Answer 24

since \(x>-1\) we know that \(1+x>0\) so the inequality stays the same if we multiply by \(1+x\) on both sides

Answer 25

doing that, we get the inequality
\[(1+x)^k(1+x)\geq (1+kx)(1+x)\] or
\[(1+x)^{k+1}\geq (1+kx)(1+x)\]

Answer 26

now the left side is just what we want
lets see what we can do with the right side
i.e. lets multiply out

Answer 27

\[(1+x)^{k+1}\geq 1+kx+x+kx^2\] which we can rewrite as
\[(1+x)^{k+1}\geq 1+(k+1)x+kx^2\]

Answer 28

now we know that \(x^2\geq 0\) and so \(kx^2\geq 0\) making a string of inequalities
\[(1+x)^{k+1}\geq 1+(k+1)x+kx^2\geq 1+(k+1)x\]
ignoring the middle term, this is exactly what we want

Answer 29

ohhhh

Answer 30

not sure if any of that was clear, or if the goal was clear, but just to restate the goal, we are trying to show that if
\[(1+x)^k\geq 1+kx\] then
\[(1+x)^{k+1}\geq 1+(k+1)x\]

Answer 31

tytytytyytyty i think i get it

Answer 32

yw
these are sometimes tricky to understand, but usually simple algebra proves them
but you can't generalize from one proof to the next
i.e. don't think they always work the same way
summations are different than inequalities