Help on induction, couldn't make both side the same value

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$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Induction g(n+1)=f(n+1)+g(n) sub n+1

(1)$$(n+1)(n+2)...(n+p)(p+1)+fracn(n+1)(n+2)...(n+p) p+1(p+1) $$
(2)
$$frac(n+1)(n+2)...(n+1+p) p+1(p+1)$$
make (1)=(2)
$$(n+p)!(p+1)+(n+p)!=(n+1+p)! $$

I then couldnt make the R.H.S==L.H.S

discrete-mathematics induction factorial recursion

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asked Jul 21 at 17:48

stevie lol

306

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up vote
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$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Induction g(n+1)=f(n+1)+g(n) sub n+1

(1)$$(n+1)(n+2)...(n+p)(p+1)+fracn(n+1)(n+2)...(n+p) p+1(p+1) $$
(2)
$$frac(n+1)(n+2)...(n+1+p) p+1(p+1)$$
make (1)=(2)
$$(n+p)!(p+1)+(n+p)!=(n+1+p)! $$

I then couldnt make the R.H.S==L.H.S

discrete-mathematics induction factorial recursion

share|cite|improve this question

asked Jul 21 at 17:48

stevie lol

306

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Induction g(n+1)=f(n+1)+g(n) sub n+1

(1)$$(n+1)(n+2)...(n+p)(p+1)+fracn(n+1)(n+2)...(n+p) p+1(p+1) $$
(2)
$$frac(n+1)(n+2)...(n+1+p) p+1(p+1)$$
make (1)=(2)
$$(n+p)!(p+1)+(n+p)!=(n+1+p)! $$

I then couldnt make the R.H.S==L.H.S

discrete-mathematics induction factorial recursion

share|cite|improve this question

asked Jul 21 at 17:48

stevie lol

306

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Induction g(n+1)=f(n+1)+g(n) sub n+1

(1)$$(n+1)(n+2)...(n+p)(p+1)+fracn(n+1)(n+2)...(n+p) p+1(p+1) $$
(2)
$$frac(n+1)(n+2)...(n+1+p) p+1(p+1)$$
make (1)=(2)
$$(n+p)!(p+1)+(n+p)!=(n+1+p)! $$

I then couldnt make the R.H.S==L.H.S

discrete-mathematics induction factorial recursion

share|cite|improve this question

asked Jul 21 at 17:48

stevie lol

306

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 17:48

stevie lol

306

asked Jul 21 at 17:48

stevie lol

306

asked Jul 21 at 17:48

stevie lol

306

306

add a comment | 

add a comment | 

2 Answers
2

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up vote
2
down vote



accepted

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)~=~fracn(n+1)(n+2)...(n+p) p+1 $$

First of all this given sum can be simplified by using factorials

$$frac(n+p)!(n-1)!~=~n(n+1)(n+2)cdots(n+p)$$

In general the term

$$(x,n)~=~x^overlinen~=~frac(x+n+1)!(x+1)!$$
is called Pochhammer symbol or rising factorial.
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~&=~fracn(n+1)(n+2)...(n+p) p+1\~&=~frac(n+p)!(n-1)!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~&=~frac(n+1)(n+2)(n+3)...(n+p+1) p+1\~&=~frac(n+p+1)!n!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~=~&sum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)n!endalign$$
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)!n!~&=~frac(n+p)!(n-1)!(p+1)~+~frac(n+p)!n!\&=~fracn(n+p)!+(p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)!n!(p+1)endalign$$

Maybe this slightly different approach to your problem can help you to get further.

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edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
  – stevie lol
  Jul 21 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
  – mrtaurho
  Jul 21 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
  – stevie lol
  Jul 21 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
  – mrtaurho
  Jul 21 at 19:09
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

I'm not sure how or why you are writing 1).

Just do the following.

Assuming that $sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$

Then $sum_k=1^n+1k(k+1)...(k+p-1) = [sum_k=1^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

$= fracn(n+1)...(n+p) p+1 + (n+1)(n+2)....(n+p-1)(n+p)$

$= [(n+1)(n+2).....(n+p)](frac np+1 + 1)$

$=[(n+1)(n+2).....(n+p)](frac np+1 + fracp+1p+1)$

$=[(n+1)(n+2).....(n+p)](frac n+p + 1p+1)$

$=frac (n+1).....(n+1 + p)p+1$

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answered Jul 21 at 19:05

fleablood

60.4k22575

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)~=~fracn(n+1)(n+2)...(n+p) p+1 $$

First of all this given sum can be simplified by using factorials

$$frac(n+p)!(n-1)!~=~n(n+1)(n+2)cdots(n+p)$$

In general the term

$$(x,n)~=~x^overlinen~=~frac(x+n+1)!(x+1)!$$
is called Pochhammer symbol or rising factorial.
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~&=~fracn(n+1)(n+2)...(n+p) p+1\~&=~frac(n+p)!(n-1)!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~&=~frac(n+1)(n+2)(n+3)...(n+p+1) p+1\~&=~frac(n+p+1)!n!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~=~&sum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)n!endalign$$
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)!n!~&=~frac(n+p)!(n-1)!(p+1)~+~frac(n+p)!n!\&=~fracn(n+p)!+(p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)!n!(p+1)endalign$$

Maybe this slightly different approach to your problem can help you to get further.

share|cite|improve this answer

edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
  – stevie lol
  Jul 21 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
  – mrtaurho
  Jul 21 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
  – stevie lol
  Jul 21 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
  – mrtaurho
  Jul 21 at 19:09
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)~=~fracn(n+1)(n+2)...(n+p) p+1 $$

First of all this given sum can be simplified by using factorials

$$frac(n+p)!(n-1)!~=~n(n+1)(n+2)cdots(n+p)$$

In general the term

$$(x,n)~=~x^overlinen~=~frac(x+n+1)!(x+1)!$$
is called Pochhammer symbol or rising factorial.
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~&=~fracn(n+1)(n+2)...(n+p) p+1\~&=~frac(n+p)!(n-1)!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~&=~frac(n+1)(n+2)(n+3)...(n+p+1) p+1\~&=~frac(n+p+1)!n!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~=~&sum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)n!endalign$$
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)!n!~&=~frac(n+p)!(n-1)!(p+1)~+~frac(n+p)!n!\&=~fracn(n+p)!+(p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)!n!(p+1)endalign$$

Maybe this slightly different approach to your problem can help you to get further.

share|cite|improve this answer

edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
  – stevie lol
  Jul 21 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
  – mrtaurho
  Jul 21 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
  – stevie lol
  Jul 21 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
  – mrtaurho
  Jul 21 at 19:09
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)~=~fracn(n+1)(n+2)...(n+p) p+1 $$

First of all this given sum can be simplified by using factorials

$$frac(n+p)!(n-1)!~=~n(n+1)(n+2)cdots(n+p)$$

In general the term

$$(x,n)~=~x^overlinen~=~frac(x+n+1)!(x+1)!$$
is called Pochhammer symbol or rising factorial.
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~&=~fracn(n+1)(n+2)...(n+p) p+1\~&=~frac(n+p)!(n-1)!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~&=~frac(n+1)(n+2)(n+3)...(n+p+1) p+1\~&=~frac(n+p+1)!n!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~=~&sum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)n!endalign$$
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)!n!~&=~frac(n+p)!(n-1)!(p+1)~+~frac(n+p)!n!\&=~fracn(n+p)!+(p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)!n!(p+1)endalign$$

Maybe this slightly different approach to your problem can help you to get further.

share|cite|improve this answer

edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

$$sum_k=1^n k(k+1)(k+2)...(k+p-1)~=~fracn(n+1)(n+2)...(n+p) p+1 $$

First of all this given sum can be simplified by using factorials

$$frac(n+p)!(n-1)!~=~n(n+1)(n+2)cdots(n+p)$$

In general the term

$$(x,n)~=~x^overlinen~=~frac(x+n+1)!(x+1)!$$
is called Pochhammer symbol or rising factorial.
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~&=~fracn(n+1)(n+2)...(n+p) p+1\~&=~frac(n+p)!(n-1)!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~&=~frac(n+1)(n+2)(n+3)...(n+p+1) p+1\~&=~frac(n+p+1)!n!(p+1)endalign$$
$$beginalignsum_k=1^n+1 frac(k+p-1)!(k-1)!~=~&sum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)n!endalign$$
$$beginalignsum_k=1^n frac(k+p-1)!(k-1)!~+~frac(n+p)!n!~&=~frac(n+p)!(n-1)!(p+1)~+~frac(n+p)!n!\&=~fracn(n+p)!+(p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)(n+p)!n!(p+1)\&=~frac(n+p+1)!n!(p+1)endalign$$

Maybe this slightly different approach to your problem can help you to get further.

share|cite|improve this answer

edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 18:18

edited Jul 21 at 18:18

edited Jul 21 at 18:18

answered Jul 21 at 18:12

mrtaurho

700219

answered Jul 21 at 18:12

mrtaurho

700219

answered Jul 21 at 18:12

mrtaurho

700219

700219

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
  – stevie lol
  Jul 21 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
  – mrtaurho
  Jul 21 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
  – stevie lol
  Jul 21 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
  – mrtaurho
  Jul 21 at 19:09
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
  – stevie lol
  Jul 21 at 18:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
  – mrtaurho
  Jul 21 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
  – stevie lol
  Jul 21 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
  – mrtaurho
  Jul 21 at 19:09
  
  

  
  
  
  

Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
– stevie lol
Jul 21 at 18:50

Do u mind explaining why when u multiply n! to (n+p)! why it renders to n(n+p)!
– stevie lol
Jul 21 at 18:50

I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
– mrtaurho
Jul 21 at 18:53

I never multiplied $(n+p)!$ ny $n!$. I used the fact that $n!=n(n-1)!$. While the final denominator of the fraction is $n!(p+1)$ I have to multiply the first fraction by $fracnn$ and the second one by $fracp+1p+1$.
– mrtaurho
Jul 21 at 18:53

Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
– stevie lol
Jul 21 at 19:06

Sry for asking a stupid question, but how did u get rid (n-1)! from the denominator and render it to n!(p+1)?
– stevie lol
Jul 21 at 19:06

There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
– mrtaurho
Jul 21 at 19:09

There is no problem in further asking :) $Big |$$frac(n+p)!(n-1)!(p+1)fracnn~=~fracn(n+p)!n(n-1)!(p+1)$ which is equal to $fracn(n+p)!n!(p+1)$ by the functional equation of the factorial $n!~=~n(n-1)!$
– mrtaurho
Jul 21 at 19:09

add a comment | 

up vote
0
down vote

I'm not sure how or why you are writing 1).

Just do the following.

Assuming that $sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$

Then $sum_k=1^n+1k(k+1)...(k+p-1) = [sum_k=1^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

$= fracn(n+1)...(n+p) p+1 + (n+1)(n+2)....(n+p-1)(n+p)$

$= [(n+1)(n+2).....(n+p)](frac np+1 + 1)$

$=[(n+1)(n+2).....(n+p)](frac np+1 + fracp+1p+1)$

$=[(n+1)(n+2).....(n+p)](frac n+p + 1p+1)$

$=frac (n+1).....(n+1 + p)p+1$

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answered Jul 21 at 19:05

fleablood

60.4k22575

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down vote

I'm not sure how or why you are writing 1).

Just do the following.

Assuming that $sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$

Then $sum_k=1^n+1k(k+1)...(k+p-1) = [sum_k=1^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

$= fracn(n+1)...(n+p) p+1 + (n+1)(n+2)....(n+p-1)(n+p)$

$= [(n+1)(n+2).....(n+p)](frac np+1 + 1)$

$=[(n+1)(n+2).....(n+p)](frac np+1 + fracp+1p+1)$

$=[(n+1)(n+2).....(n+p)](frac n+p + 1p+1)$

$=frac (n+1).....(n+1 + p)p+1$

share|cite|improve this answer

answered Jul 21 at 19:05

fleablood

60.4k22575

add a comment | 

up vote
0
down vote

up vote
0
down vote

I'm not sure how or why you are writing 1).

Just do the following.

Assuming that $sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$

Then $sum_k=1^n+1k(k+1)...(k+p-1) = [sum_k=1^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

$= fracn(n+1)...(n+p) p+1 + (n+1)(n+2)....(n+p-1)(n+p)$

$= [(n+1)(n+2).....(n+p)](frac np+1 + 1)$

$=[(n+1)(n+2).....(n+p)](frac np+1 + fracp+1p+1)$

$=[(n+1)(n+2).....(n+p)](frac n+p + 1p+1)$

$=frac (n+1).....(n+1 + p)p+1$

share|cite|improve this answer

answered Jul 21 at 19:05

fleablood

60.4k22575

I'm not sure how or why you are writing 1).

Just do the following.

Assuming that $sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$

Then $sum_k=1^n+1k(k+1)...(k+p-1) = [sum_k=1^n k(k+1)...(k+p-1)]+ (n+1)((n+1)+1)...((n+1)+p-1)$

$= fracn(n+1)...(n+p) p+1 + (n+1)(n+2)....(n+p-1)(n+p)$

$= [(n+1)(n+2).....(n+p)](frac np+1 + 1)$

$=[(n+1)(n+2).....(n+p)](frac np+1 + fracp+1p+1)$

$=[(n+1)(n+2).....(n+p)](frac n+p + 1p+1)$

$=frac (n+1).....(n+1 + p)p+1$

share|cite|improve this answer

answered Jul 21 at 19:05

fleablood

60.4k22575

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 19:05

fleablood

60.4k22575

answered Jul 21 at 19:05

fleablood

60.4k22575

answered Jul 21 at 19:05

fleablood

60.4k22575

60.4k22575

add a comment | 

add a comment | 

 

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