For every real \(t>-1\), and every \(n \in N\), we have
\[(1+t)^n \ge 1+nt, (1+t)^n \ge 1+nt+\frac{1}{2}n(n-1)t^2;\]when \(t \neq 0\), the inequalities are strict for \(n>1\),\(n>2\), respectively.

- anonymous

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- anonymous

@amistre64 i opened this one

- anonymous

i don't quite understand what I'm proving

- amistre64

the indication that t not= 0 suggests that we have to divide by t, to get to whatever we have to get to ...

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## More answers

- anonymous

ok

- anonymous

i'm still kind of confused as to where to start

- amistre64

im thinking its 2 problems kind of combined

- amistre64

and you have to induce it to n, and n+1

- anonymous

it's actually under "Lemma" and it tells me to prove this lemma

- anonymous

ok so i'll prove it by induction the way we did it before?

- amistre64

\[(1+t)^n \ge 1+nt~:~for~n>1\]
\[(1+t)^n \ge 1+nt+\frac{1}{2}n(n-1)t^2~:~for~n>2\]
yeah, its pretty much the same concept

- amistre64

let n=2 for the first one, and prove it; then set it up for an n+1 run

- anonymous

ok and then let n=3 for the second one?

- amistre64

we know that (1+x)^n is equal to some binomial exapnsion ... so that might be needed as well

- anonymous

ok

- amistre64

im thinking this plays into the proof that we just covered :)

- amistre64

the right sides are expansions, at least partial, and we can use the expansion of the binomial for the first few terms is a thought

- anonymous

i'm kind of lost... should i start with the left side first? it looks less complicated

- amistre64

im thinking we can replace the left sides by the summation stuff, and pull out a few terms for comparison

- anonymous

i was using playtex and it crashed on me

- amistre64

\[(1+t)^n \ge 1+nt\]
\[\sum_{k=0}^{n}\binom nk x^k \ge 1+nt~:~let~n=2\]
\[\sum_{k=0}^{2}\binom 2k x^k \ge 1+2t\]
\[1+2t+t^2\ge 1+2t\]for t\(\ne\)0
\[t^2\ge0\]

- amistre64

since t not=0, the equal bar underneath is moot

- amistre64

we know the left summation is good for an n+1 .... or whatever variable you want to redefine it as

- amistre64

multiply both sides by (1+t) like we did before, and it should be a cake walk

- anonymous

ok i got that part. but what i don't get is what does that prove?

- anonymous

i got the t^2>0 part. i don't know what to do with that information though

- amistre64

it proves that it is strictly greater than the right side for n=2, we need to show that it is true for 2+1, for 2+1+1, for 2+1+1+....+1

- anonymous

ok

- amistre64

thats just the basis step, its true for n=2; let n=s i guess; and show that its true for some s+1

- amistre64

n=\(\mu\) if you wan to get all greecey

- anonymous

what should i expect for the result of n=s then? that (1+t)^s>=1+st?

- amistre64

we should that:\[(1+t)\sum\binom \mu kx^k = \sum \binom{\mu+1}{k}x^k\]

- amistre64

we *showed .... that is

- anonymous

ok i'm condused. sorry i'm kind of slow

- amistre64

multiplying both sides by (1+t) is what we need to demonstrate right?

- amistre64

crashed again ...

- anonymous

yes
from the last proof right?

- amistre64

\[(1+t)^n>1+nt\]
\[(1+t)(1+t)^n>(1+nt)(1+t)\]
\[(1+t)^{n+1}>(1+nt)(1+t)\]
\[\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]

- amistre64

drops some expansions out the left side ... what do start to get?

- anonymous

ok i understand this part

- anonymous

do we just ignore the equal under the > tho?

- amistre64

since t has a domain of (-1,0), 1 - fraction is never negative so we aint got to bother worrying about a sign flip either

- anonymous

ok!

- amistre64

we know that its stricly > for n=2, we want to show that it is strictly true for numbers bigger than 2 as well

- anonymous

i see

- anonymous

how do i get the st^2 to go away on the right?

- amistre64

\[\binom{\mu+1}{0}+\binom{\mu+1}{1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]
\[1+{\mu+1}t+\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~1+(\mu+1)t+\mu t^2\]
\[\binom{\mu+1}{2}t^2+\sum_{k=0}^{\mu+1}\binom{\mu+1}{k}t^k~>~\mu t^2\]
whats the right up for the (u+1 2), and how does it relate to u?

- amistre64

*the write up

- anonymous

(u+1,2)=(u+1)!/(2*(u-1)!) right?

- amistre64

\[\frac{(\mu+1)!}{(\mu+1-2)!}\]
\[\frac{(\mu+1)!}{(\mu-1)!}\]
\[\frac{(\mu+1)(\mu)\cancel{(\mu-1)!}}{\cancel{(\mu-1)!}}\]
is...
\[\mu(\mu+1)>\mu\]

- anonymous

i thought there would be a 2 at the bottom because of the k!(n-k)! at the bottom?

- amistre64

\[\binom {n}{k}=\frac{n!}{(n-k)!}\]
\[\binom {\mu+1}{2}=\frac{(\mu+1)!}{(\mu+1-2)!}\]

- amistre64

\[\mu(\mu+1)>\mu\]
\[\mu(\mu+1)-\mu>0\]
\[\mu(\mu+1-1)>0\]
\[\mu^2>0~:~for~all~\mu>1\]

- anonymous

\[\left(\begin{matrix}n \\ k\end{matrix}\right)\] is actually \[\frac{n!}{k!(n-k)!}\]in my textbook

- amistre64

.... hmm, you may be right about that :) write it up and see what we get ... i got too many things rolling around in me head to sort them

- amistre64

\[\frac12\mu(\mu+1)>\mu\]happier :)

- anonymous

ok i get that and what should i do next?

- anonymous

i got \[s^2-s>0\]

- amistre64

or written another way:
s^2 > s isnt this true for all s>2 ?

- anonymous

yes so that means we prove the first part by induction?

- amistre64

it shows that the left side is bigger than the right side:
(Bigger) t^2 + (more) > (smaller) t^2

- amistre64

yes, the first part is proved

- anonymous

yay ok so i think the second thing is actually the same because 1/2(n)(n+1) is actually (n+1,2) so i think we cancel that out so i can use the to prove k=3

- amistre64

at least I believe it is

- amistre64

the second one would follow the same concepts yes

- anonymous

thank you so much!

- amistre64

youre welcome :)

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