Proving (1+n)^n≥ 5/2* n^n- 1/2* n^(n-1) for n≥2 using Binomial Theorem

[rohan03]

Homework Statement

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

Homework Equations

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

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]

Homework Statement

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

Homework Equations

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

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 !

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]

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

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]

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

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]

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

$n$ 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]

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

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

 

[Infinitum]

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.

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

$$n! = n\cdot (n-1) \cdot (n-2) ... 2\cdot 1$$

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

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

https://www.physicsforums.com/library.php?do=view_item&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 don't that correct??

 

[LCKurtz]

so say i was doing k = n-1

n(n-1)/n(n-1)(0)! = n(n-1)/0! = n(n-1)? have I don't that correct??

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

 

[fireychariot]

thank you for your help so far.

I want to see if i have expanded this right then. Definetly need more practice on this.

(1+n)^n = 1^n + 1^n + n^n-1 + 1/2n^n-2 + $\cdots$

 

[Infinitum]

thank you for your help so far.

I want to see if i have expanded this right then. Definetly need more practice on this.

(1+n)^n = 1^n + 1^n + n^n-1 + 1/2n^n-2 + $\cdots$

Use parenthesis! I can't figure out the equation completely, but it is surely incorrect as you won't get 1ⁿ twice. The binomial expansion is given as,

$$(a+b)^n = \binom{n}{0} a^{n}b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + ...$$

Keeping in mind what LCKurtz said, what do you get if you put a=1, b=n, and n=n in the above equation?

Also, did you check out the link in my previous post??

 

[fireychariot]

$\frac{n!}{0!(n-0)!}$1$^{n}$ + $\frac{n!}{(n!)(n-n)!}$n$^{n}$ + $\frac{n!}{(n-1)!(n-(n-1))}$n$^{n-1}$+$\frac{n!}{(n-2)!(n-(n-2))}$n$^{n-2}$ + $\cdots$

= 1 + n$^{n}$ + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?

 

[Infinitum]

$\frac{n!}{0!(n-0)!}$1$^{n}$ + $\frac{n!}{(n!)(n-n)!}$n$^{n}$ + $\frac{n!}{(n-1)!(n-(n-1))}$n$^{n-1}$+$\frac{n!}{(n-2)!(n-(n-2))}$n$^{n-2}$ + $\cdots$

= (1 + n)$^{n}$ + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?

Yes, that's correct, and probably what you are looking for.

 

[fireychariot]

$\frac{n!}{0!(n-0)!}$1$^{n}$ + $\frac{n!}{(n!)(n-n)!}$n$^{n}$ + $\frac{n!}{(n-1)!(n-(n-1))}$n$^{n-1}$+$\frac{n!}{(n-2)!(n-(n-2))}$n$^{n-2}$ + $\cdots$

= 1 + n$^{n}$ + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?

What would the coefficients of n-1 and n-2 be then?

 

[Infinitum]

What would the coefficients of n-1 and n-2 be then?

You mean the coefficients of n^n-1?

From your previous post..

$$\frac{n!}{(n-1)!(n-(n-1))!}n^{n-1}$$

 

[fireychariot]

Yes that's correct many thanks