Cool one

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Cool one

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

17 posts

#1 h Yesterday at 9:09 PM

Y by

Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

239 posts

vsarg

#2 h Yesterday at 10:11 PM

Y by

Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

:blush:

This post has been edited 1 time. Last edited by vsarg, Yesterday at 10:12 PM
Reason: Many mistake for me. I edit to ffix it

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1096 posts

#3 h Yesterday at 10:32 PM

Y by

vsarg wrote:

Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

:blush:

Not valid you didnt prove.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

17 posts

#4 h Yesterday at 10:42 PM

Y by

vsarg wrote:

Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

:blush:

EXACTLY WHY IS IT GREATER THAT $(n+1) \cdot b^n$ ?

This post has been edited 1 time. Last edited by MTA_2024, Yesterday at 10:42 PM
Reason: Miswriting

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

no_room_for_error

316 posts

no_room_for_error

#5 h Yesterday at 11:55 PM

Y by

EaZ_Shadow wrote:

vsarg wrote:

Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

:blush:

Not valid you didnt prove.

His proof is valid (and imo explained thoroughly enough).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ohiorizzler1434

691 posts

ohiorizzler1434

#6 h Yesterday at 11:59 PM

Y by

MTA_2024 wrote:

Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n$

bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!

I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!

This post has been edited 1 time. Last edited by ohiorizzler1434, Yesterday at 11:59 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

17 posts

#7 h Today at 12:19 AM

Y by

Sorry I'm dumb, found it out myself a bit later.
I was stuck on a problem till I reached right here. Thought I was still a long way through, before realising what is the fudging problem quoting at the very beginning. $a>b$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a