Cool one

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MTA_2024

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MTA_2024

#1 h Saturday at 9:09 PM

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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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vsarg

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vsarg

#2 h Saturday at 10:11 PM

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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:

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

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EaZ_Shadow

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EaZ_Shadow

#3 h Saturday at 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:

Not valid you didnt prove.

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MTA_2024

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MTA_2024

#4 h Saturday at 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:

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

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

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no_room_for_error

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no_room_for_error

#5 h Saturday at 11:55 PM

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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:

Not valid you didnt prove.

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

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ohiorizzler1434

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ohiorizzler1434

#6 h Saturday at 11:59 PM

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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, Saturday at 11:59 PM

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MTA_2024

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MTA_2024

#7 h Yesterday at 12:19 AM

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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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