binomial theorem to prove

rohan03

1. The problem statement, all variables and given/known data

(1+n)^n≥ 5/2* n^n- 1/2* n^(n-1) for n≥2

2. Relevant equations
i know I have to use this formula
(1+x)^n=1+nx/1!+(n(n-1) x^2)/2!+⋯

3. The attempt at a solution

And you take x=n from my original inequality but after that I have no clue
(1+n)^n=1+n/1! n+(n(n-1) n^2)/2!+⋯
but it seems very complicated !!

LCKurtz

Since n is a positive integer, that binomial expansion has finitely many terms. And since ##n\ge 2## the expansion has at least 3 terms. Try keeping just the last three terms (which you didn't write down) because they are the ones with the higher powers of n.

rohan03

Will do that and write my results here. What I don't know is how to get last terms power to n-1 ?

rohan03

I'll be doing this tomorrow so please check ur post.

rohan03

(1+n)^n=1+n/1 n+(n(n-1) n^2)/2!+(n(n-1) 〖(n-2)n〗^3)/3!…

LCKurtz

Do you know the binomial expansion for ##(a+b)^n##? It doesn't end with a "...". You need to look at the last three terms, not the first terms.

rohan03

if I use just the last three term I do get the result- so is it ok for me to just use last three terms ignoring first few terms- if yes than I have resolved so simple problem- which looked very complicated in the first place- Thank you

LCKurtz

Yes. That is why there is a ##\ge## sign in the statement of the problem. Leaving out the other terms makes it smaller.

rohan03

thank you- this forum is blessing to someone relatively new to pure maths.

twinplums

I'm confused.

Which binomial expansion should I be using for this question and why?

Is it binomial expansion for (a+b)^n or (1+x)^n

fireychariot

what do you mean last 3 terms? if its infinite it wouldn't have any last term?

LCKurtz

##n## is a positive integer so the binomial expansion is finite:$$
(1+n)^n =\sum_{k=0}^n \binom n k 1^{n-k} n^k$$The last three terms are for ##k=n,n-1,n-2##.

fireychariot

Thanks for that. I'm having a blonde moment though when working out the coefficients. I know you use n!/k!(n-k)! However I am little confused when I input for say k = n-1 I get n!/(n-1)!(n-n-1)! How do I expand on this? Many thanks.

Jorriss

They're from the same formula. Set a=1, b=x.

Infinitum

A factorial for natural numbers is the product of all consecutive integers from 1 upto n. That is,

[tex]n! = n\cdot (n-1) \cdot (n-2) ........ 2\cdot 1[/tex]

Use this definition in both the numerator and denominator of the expression. What do you get?



http://physicsforums.com/library.php...tem&itemid=869

fireychariot

so say i was doing k = n-1

n(n-1)/n(n-1)(0)! = n(n-1)/0! = n(n-1)? have I dont that correct??

LCKurtz

No.$$
\binom n {n-1}=\frac {n!}{(n-1)!(n-(n-1))!}=\frac {n!}{(n-1)!(n-n+1)!}
=n$$