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even number - binomial

Property of Binomial Coefficientappling the binomial theoremBinomial Expansion.A binomial random number generating algorithm that works when $ n times p $ is very smallq-binomial IdentityAsymptotic bound for products of even binomial coefficientsEvaluating sum of binomial coefficientsSummation of binomial coefficients up to $r$ terms.Binomial summationsHow to prove binomial coefficient $ 2^n choose k $ is even number?

-1

$begingroup$

$$(1+sqrt 2)^n+(1-sqrt2)^n$$
if $$ n in N $$
It's always an even number.

I tried to solve for the binomial, but I could not, any idea to be able to proceed
$$(a+b)^n=sum_k=0^nn choose k a^n-kb^k$$

binomial-coefficients

share|cite|improve this question

asked Mar 29 at 15:08

MonicaMonica

215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proceed by induction
  $endgroup$
  – Jakobian
  Mar 29 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have to do without induction, help
  $endgroup$
  – Monica
  Mar 29 at 15:14
  

  
  
  
  

add a comment | 

-1

$begingroup$

$$(1+sqrt 2)^n+(1-sqrt2)^n$$
if $$ n in N $$
It's always an even number.

I tried to solve for the binomial, but I could not, any idea to be able to proceed
$$(a+b)^n=sum_k=0^nn choose k a^n-kb^k$$

binomial-coefficients

share|cite|improve this question

asked Mar 29 at 15:08

MonicaMonica

215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proceed by induction
  $endgroup$
  – Jakobian
  Mar 29 at 15:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have to do without induction, help
  $endgroup$
  – Monica
  Mar 29 at 15:14
  

  
  
  
  

add a comment | 

$begingroup$

$$(1+sqrt 2)^n+(1-sqrt2)^n$$
if $$ n in N $$
It's always an even number.

I tried to solve for the binomial, but I could not, any idea to be able to proceed
$$(a+b)^n=sum_k=0^nn choose k a^n-kb^k$$

binomial-coefficients

share|cite|improve this question

asked Mar 29 at 15:08

MonicaMonica

215

$endgroup$

share|cite|improve this question

asked Mar 29 at 15:08

MonicaMonica

215

share|cite|improve this question

asked Mar 29 at 15:08

MonicaMonica

215

asked Mar 29 at 15:08

MonicaMonica

215

asked Mar 29 at 15:08

MonicaMonica

215

3 Answers
3

active

oldest

votes

2

$begingroup$

Applying the binomial expansion you gave,

$$(1+sqrt2)^n=sum_k=0^nn choose k sqrt2^k$$

$$(1-sqrt2)^n=sum_k=0^nn choose k (-sqrt2)^k$$

In their sum, terms of odd $k$ cancel, and terms of even $k$ are duplicated, so the sum is $$2sum_j=0^lfloorfrac n2rfloorn choose 2j sqrt2^2j,$$

which is clearly an even integer, because it's $2$ times an integer --

binomial coefficients and even non-negative powers of $sqrt2$ are integers -- no induction needed.

share|cite|improve this answer

edited Mar 31 at 5:08

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I used the fact that $(-1)^k=1$ if $k$ is even and $-1$ if $k$ is odd
  $endgroup$
  – J. W. Tanner
  Mar 29 at 15:42
  

  
  
  
  

add a comment | 

1

$begingroup$

For all $n in mathbbN$, let $u_n = left( 1 +sqrt2 right)^n + left(1- sqrt2 right)^n$.

First, for all $n$, one has $u_n+2=2u_n+1 + u_n$ : indeed
$$u_n+2 = left( 1 +sqrt2 right)^n+2 + left(1- sqrt2 right)^n+2 = left( 1 +sqrt2 right)^2left( 1 +sqrt2 right)^n + left( 1 -sqrt2 right)^2left(1- sqrt2 right)^n $$

$$= (1+2(sqrt2+1))left( 1 +sqrt2 right)^n + left( 1 -2(sqrt2+1) right)left(1- sqrt2 right)^n$$

$$= left( 1 +sqrt2 right)^n + 2 left( 1 +sqrt2 right)^n+1 + left(1- sqrt2 right)^n - 2left(1- sqrt2 right)^n+1$$

So you get
$$u_n+2=2u_n+1 + u_n$$

Now, a small induction shows you that $(u_n)$ is even for each $n$.

share|cite|improve this answer

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

0

$begingroup$

You can restore the difference equation with roots $1-sqrt2$ and $1+sqrt2$:
$$x^2-2x-1=0.$$
Hence:
$$a_n=2a_n-1+a_n-2, a_1=2,a_2=6,$$
which has a solution:
$$a_n=(1-sqrt2)^n+(1+sqrtn)^n.$$
Obviously, it has all terms even.

share|cite|improve this answer

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You must have meant $sqrt 2$ where you wrote $sqrt n$
  $endgroup$
  – J. W. Tanner
  Mar 29 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.W.Tanner, yes, indeed, let your note be erratum.
  $endgroup$
  – farruhota
  Mar 29 at 19:59
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

Applying the binomial expansion you gave,

$$(1+sqrt2)^n=sum_k=0^nn choose k sqrt2^k$$

$$(1-sqrt2)^n=sum_k=0^nn choose k (-sqrt2)^k$$

In their sum, terms of odd $k$ cancel, and terms of even $k$ are duplicated, so the sum is $$2sum_j=0^lfloorfrac n2rfloorn choose 2j sqrt2^2j,$$

which is clearly an even integer, because it's $2$ times an integer --

binomial coefficients and even non-negative powers of $sqrt2$ are integers -- no induction needed.

share|cite|improve this answer

edited Mar 31 at 5:08

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I used the fact that $(-1)^k=1$ if $k$ is even and $-1$ if $k$ is odd
  $endgroup$
  – J. W. Tanner
  Mar 29 at 15:42
  

  
  
  
  

add a comment | 

2

$begingroup$

Applying the binomial expansion you gave,

$$(1+sqrt2)^n=sum_k=0^nn choose k sqrt2^k$$

$$(1-sqrt2)^n=sum_k=0^nn choose k (-sqrt2)^k$$

In their sum, terms of odd $k$ cancel, and terms of even $k$ are duplicated, so the sum is $$2sum_j=0^lfloorfrac n2rfloorn choose 2j sqrt2^2j,$$

which is clearly an even integer, because it's $2$ times an integer --

binomial coefficients and even non-negative powers of $sqrt2$ are integers -- no induction needed.

share|cite|improve this answer

edited Mar 31 at 5:08

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I used the fact that $(-1)^k=1$ if $k$ is even and $-1$ if $k$ is odd
  $endgroup$
  – J. W. Tanner
  Mar 29 at 15:42
  

  
  
  
  

add a comment | 

$begingroup$

Applying the binomial expansion you gave,

$$(1+sqrt2)^n=sum_k=0^nn choose k sqrt2^k$$

$$(1-sqrt2)^n=sum_k=0^nn choose k (-sqrt2)^k$$

In their sum, terms of odd $k$ cancel, and terms of even $k$ are duplicated, so the sum is $$2sum_j=0^lfloorfrac n2rfloorn choose 2j sqrt2^2j,$$

which is clearly an even integer, because it's $2$ times an integer --

binomial coefficients and even non-negative powers of $sqrt2$ are integers -- no induction needed.

share|cite|improve this answer

edited Mar 31 at 5:08

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

$endgroup$

share|cite|improve this answer

edited Mar 31 at 5:08

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

answered Mar 29 at 15:14

J. W. TannerJ. W. Tanner

4,4891320

1

$begingroup$

For all $n in mathbbN$, let $u_n = left( 1 +sqrt2 right)^n + left(1- sqrt2 right)^n$.

First, for all $n$, one has $u_n+2=2u_n+1 + u_n$ : indeed
$$u_n+2 = left( 1 +sqrt2 right)^n+2 + left(1- sqrt2 right)^n+2 = left( 1 +sqrt2 right)^2left( 1 +sqrt2 right)^n + left( 1 -sqrt2 right)^2left(1- sqrt2 right)^n $$

$$= (1+2(sqrt2+1))left( 1 +sqrt2 right)^n + left( 1 -2(sqrt2+1) right)left(1- sqrt2 right)^n$$

$$= left( 1 +sqrt2 right)^n + 2 left( 1 +sqrt2 right)^n+1 + left(1- sqrt2 right)^n - 2left(1- sqrt2 right)^n+1$$

So you get
$$u_n+2=2u_n+1 + u_n$$

Now, a small induction shows you that $(u_n)$ is even for each $n$.

share|cite|improve this answer

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

1

$begingroup$

For all $n in mathbbN$, let $u_n = left( 1 +sqrt2 right)^n + left(1- sqrt2 right)^n$.

First, for all $n$, one has $u_n+2=2u_n+1 + u_n$ : indeed
$$u_n+2 = left( 1 +sqrt2 right)^n+2 + left(1- sqrt2 right)^n+2 = left( 1 +sqrt2 right)^2left( 1 +sqrt2 right)^n + left( 1 -sqrt2 right)^2left(1- sqrt2 right)^n $$

$$= (1+2(sqrt2+1))left( 1 +sqrt2 right)^n + left( 1 -2(sqrt2+1) right)left(1- sqrt2 right)^n$$

$$= left( 1 +sqrt2 right)^n + 2 left( 1 +sqrt2 right)^n+1 + left(1- sqrt2 right)^n - 2left(1- sqrt2 right)^n+1$$

So you get
$$u_n+2=2u_n+1 + u_n$$

Now, a small induction shows you that $(u_n)$ is even for each $n$.

share|cite|improve this answer

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

$begingroup$

For all $n in mathbbN$, let $u_n = left( 1 +sqrt2 right)^n + left(1- sqrt2 right)^n$.

First, for all $n$, one has $u_n+2=2u_n+1 + u_n$ : indeed
$$u_n+2 = left( 1 +sqrt2 right)^n+2 + left(1- sqrt2 right)^n+2 = left( 1 +sqrt2 right)^2left( 1 +sqrt2 right)^n + left( 1 -sqrt2 right)^2left(1- sqrt2 right)^n $$

$$= (1+2(sqrt2+1))left( 1 +sqrt2 right)^n + left( 1 -2(sqrt2+1) right)left(1- sqrt2 right)^n$$

$$= left( 1 +sqrt2 right)^n + 2 left( 1 +sqrt2 right)^n+1 + left(1- sqrt2 right)^n - 2left(1- sqrt2 right)^n+1$$

So you get
$$u_n+2=2u_n+1 + u_n$$

Now, a small induction shows you that $(u_n)$ is even for each $n$.

share|cite|improve this answer

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

share|cite|improve this answer

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

answered Mar 29 at 15:21

TheSilverDoeTheSilverDoe

5,433216

0

$begingroup$

You can restore the difference equation with roots $1-sqrt2$ and $1+sqrt2$:
$$x^2-2x-1=0.$$
Hence:
$$a_n=2a_n-1+a_n-2, a_1=2,a_2=6,$$
which has a solution:
$$a_n=(1-sqrt2)^n+(1+sqrtn)^n.$$
Obviously, it has all terms even.

share|cite|improve this answer

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You must have meant $sqrt 2$ where you wrote $sqrt n$
  $endgroup$
  – J. W. Tanner
  Mar 29 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.W.Tanner, yes, indeed, let your note be erratum.
  $endgroup$
  – farruhota
  Mar 29 at 19:59
  

  
  
  
  

add a comment | 

0

$begingroup$

You can restore the difference equation with roots $1-sqrt2$ and $1+sqrt2$:
$$x^2-2x-1=0.$$
Hence:
$$a_n=2a_n-1+a_n-2, a_1=2,a_2=6,$$
which has a solution:
$$a_n=(1-sqrt2)^n+(1+sqrtn)^n.$$
Obviously, it has all terms even.

share|cite|improve this answer

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You must have meant $sqrt 2$ where you wrote $sqrt n$
  $endgroup$
  – J. W. Tanner
  Mar 29 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.W.Tanner, yes, indeed, let your note be erratum.
  $endgroup$
  – farruhota
  Mar 29 at 19:59
  

  
  
  
  

add a comment | 

$begingroup$

You can restore the difference equation with roots $1-sqrt2$ and $1+sqrt2$:
$$x^2-2x-1=0.$$
Hence:
$$a_n=2a_n-1+a_n-2, a_1=2,a_2=6,$$
which has a solution:
$$a_n=(1-sqrt2)^n+(1+sqrtn)^n.$$
Obviously, it has all terms even.

share|cite|improve this answer

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

$endgroup$

share|cite|improve this answer

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842

answered Mar 29 at 15:22

farruhotafarruhota

21.8k2842