Yahoo Answers: Answers and Comments for Prove that 7(3^(2n)2^n) for every nonnegative integer n.? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
From Anonymous
enSG
Fri, 01 Oct 2010 12:08:15 +0000
3
Yahoo Answers: Answers and Comments for Prove that 7(3^(2n)2^n) for every nonnegative integer n.? [Mathematics]
292
38
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://s.yimg.com/zz/combo?images/emaillogosg.png

From Todd: I don't remember the details of some of th...
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Fri, 01 Oct 2010 12:18:46 +0000
I don't remember the details of some of these theorems, but what I write below is true.
3^(2n) = (3^2)^n = 9^n. So we have 9^n  2^n. Because 9 = 2+7, 9^n (mod 7) = 2^n (mod 7). Thus, we have 9^n  2^n (mod 7) = 2^n  2^n (mod 7) = 0 (mod 7). Therefore, it is divisible by 7. Note: The "=" signs should be the threelined equivalent signs.

From palomares: i think of you propose n^3 + 2n Induction will...
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Sat, 03 Dec 2016 15:51:01 +0000
i think of you propose n^3 + 2n Induction will sparkling up this. For base case n = a million, a million^3 + 2 = 3 is divisible via 3. anticipate it incredibly is actual for all n = ok > ok^3 + 2k is divisible via 3. enable ok^3 + 2k = 3m (m is an integer) Proving for n = ok + a million (ok + a million)^3 + 2(ok + a million) = ok^3 + a million + 3k^2 + 3k + 2k + 2 = (ok^3 + 2k) + 3k^2 + 3k + 3 = 3m + 3k^2 + 3k + 3 = 3(m + ok^2 + ok + 3) that's divisible via 3 considering that n = ok is actual implies n = ok + a million is actual, and since the backside case is actual to boot, n^3 + 2n is divisible via 3 for all n ? Z

From Kusmosis: An easy proof:
3^(2n)  2^n = 9^n  2^n
...
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Sun, 03 Oct 2010 09:35:09 +0000
An easy proof:
3^(2n)  2^n = 9^n  2^n
Lemma: (ab)  (a^n  b^n) [THIS IS actually a very important fact]
Proof: Consider the expression a^(n1) * b + a^(n2) * b^2 + ... a^2 * b^(n2) + a * b^(n1)
Write this in terms of a geometric series (it may take you a while to see this)
(a^(n1) * b)((b/a)^(n1)  1)

(b/a)  1
after loads of simplification, this becomes [ab(a^n  b^n)]/(ab)
however since we know that expression is an integer, (ab)  (a^n  b^n)
hence, (92)  (9^n  2^n) ==> 7  (3^(2n)  2^n)

From δοτζο: 7  (3²ⁿ  2ⁿ) ⇔ 3²ⁿ  2ⁿ = 7u ⇔ 9ⁿ  2ⁿ = 7u,...
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Fri, 01 Oct 2010 12:58:37 +0000
7  (3²ⁿ  2ⁿ) ⇔ 3²ⁿ  2ⁿ = 7u ⇔ 9ⁿ  2ⁿ = 7u, where u ∈ ℤ
9ⁿ = (2+7)ⁿ = C(n,0)(2ⁿ)(7⁰) + C(n,1)(2ⁿ⁻¹)(7) + C(n,2)(2ⁿ⁻²)(7²) + ... + C(n,n1)(2¹)(7ⁿ⁻¹) + C(n,n)(2⁰)(7ⁿ)
9ⁿ  2ⁿ = 2ⁿ + C(n,1)(2ⁿ⁻¹)(7) + C(n,2)(2ⁿ⁻²)(7²) + ... + C(n,n1)(2¹)(7ⁿ⁻¹ + C(n,n)(2⁰)(7ⁿ)  2ⁿ
= C(n,1)(2ⁿ⁻¹)(7) + C(n,2)(2ⁿ⁻²)(7²) + ... + C(n,n1)(2¹)(7ⁿ⁻¹) + C(n,n)(2⁰)(7ⁿ)
= 7(C(n,1)(2ⁿ⁻¹) + C(n,2)(2ⁿ⁻²)(7¹) + ... + C(n,n1)(2¹)(7ⁿ⁻²) + C(n,n)(2⁰)(7ⁿ⁻¹))
= 7u
u = (C(n,1)(2ⁿ⁻¹) + C(n,2)(2ⁿ⁻²)(7¹) + ... + C(n,n1)(2¹)(7ⁿ⁻²) + C(n,n)(2⁰)(7ⁿ⁻¹))
■

From dophse: what needs to be proved is not definitive
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Fri, 01 Oct 2010 12:14:22 +0000
what needs to be proved is not definitive

From The Grannyator: what the heck is that  sign?
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
https://sg.answers.yahoo.com/question/index?qid=20101001120815AAtwoKi
Fri, 01 Oct 2010 12:13:05 +0000
what the heck is that  sign?