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

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24 posts

#1 h Mar 15, 2025, 9:09 PM • 1 Y

Y by byron-aj-tom

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

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250 posts

vsarg

#2 h Mar 15, 2025, 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, Mar 15, 2025, 10:12 PM
Reason: Many mistake for me. I edit to ffix it

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1154 posts

#3 h Mar 15, 2025, 10:32 PM

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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.

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#4 h Mar 15, 2025, 10:42 PM

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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, Mar 15, 2025, 10:42 PM
Reason: Miswriting

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no_room_for_error

326 posts

no_room_for_error

#5 h Mar 15, 2025, 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).

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ohiorizzler1434

744 posts

ohiorizzler1434

#6 h Mar 15, 2025, 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, Mar 15, 2025, 11:59 PM

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#7 h Mar 16, 2025, 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. $a>b$

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#8 h Mar 20, 2025, 8:52 AM

Y by

This is the natural solution above! I have an idea that is not only valid for natural numbers $n$ ! Let's see! Consider the function $f(x)=x^{n+1}$ on the interval $[a;b]$ . If we apply Lagrange's finite growth theorem, then there is a $c$ in the interval $(a;b)$ for which $\frac{f(b)-f(a)}{b-a}=(n+1){{c}^{n}}$ But since $a<c<b$ we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers $n$ !

This post has been edited 1 time. Last edited by invisibleman, Mar 20, 2025, 8:53 AM

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ohiorizzler1434

744 posts

ohiorizzler1434

#9 h Mar 20, 2025, 9:58 AM

Y by

what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.

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#10 h Mar 20, 2025, 10:56 AM

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ohiorizzler1434 wrote:

what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.

Dear friend! Maybe I didn't express myself clearly. I meant this:
https://en.wikipedia.org/wiki/Mean_value_theorem

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250 posts

vsarg

#11 h Mar 29, 2025, 6:09 PM

Y by

ohiorizzler1434 wrote:

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!

Very correct Ohifrizzler thank yod for baking me up from meanies like Eazy. In Kentucky we would :moose:

:moose:

horse race them to seeee whos the better one.

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41393 posts

sqing

#12 h Mar 30, 2025, 2:42 AM

Y by

MTA_2024 wrote:

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

https://math.stackexchange.com/questions/71285/series-apostol-calculus-vol-i-section-10-20-24?noredirect=1
https://math.stackexchange.com/questions/3890494/how-could-someone-conceive-of-using-this-inequality-for-this-proof?noredirect=1

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