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Mathematical induction - recursion

Prove the inequality $n! geq 2^n$ by inductionReason behind particular induction stepMathematical induction for inequalities with a constant at the right sideApplications of mathematical inductionMathematical induction--When it can and can't be usedHow have they done the algebra here?In-Depth Explanation of How to Do Mathematical Induction Over the Set $mathbb{R}$ of All Real Numbers?Interpretation of 2 proofs involving limits at infinity and mathematical inductionShow that an inequality is true using mathematical induction and the mean value theoremOn Assumptions In Induction

0

$begingroup$

I am struggling with that exercise... i don't understand a step in the answer. Maybe somebody Can explain in here or at least give me a hint what's going on there.

$a_{n} = frac{7*(n+2)}{2*n} *a_{n-1}$ ,$a_{1} = 14$

So it's...

$a_{2} = frac{7*4}{4} * a_{1} = frac{7*4 * 14}{4} = frac{2* 7^{2}*4 }{4}$

$a_{3} = frac{7*5}{6} * a_{2} = frac{2* 7^{3}*4*5 }{4*6}$

And here is the problem...
a4 equals:

$a_{4} = frac{7*6}{8} * a_{3} = frac{2* 7^{4}*4*5*6 }{4*6*8}$

We are multiplying numerator and denominator with 2*3 to get factorial 6! on numerator. I can't understand how the
$(1*2*3)^2$ got on denominator and how after that we transformed it to $3! * 4!$. Next steps are quite easy and the mathematical induction is fully understandable for me.

$a_{4} = frac{2* 7^{4}*6! }{ (1*2*3)^{2} * 2 *3 *4 } = frac{2* 7^{4}*6! }{ 3! 4! }$

Full answer of the exercise: JPG image

induction recursion

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asked Mar 20 at 12:14

Jim.DJim.D

31

$endgroup$

add a comment | 

0

$begingroup$

I am struggling with that exercise... i don't understand a step in the answer. Maybe somebody Can explain in here or at least give me a hint what's going on there.

$a_{n} = frac{7*(n+2)}{2*n} *a_{n-1}$ ,$a_{1} = 14$

So it's...

$a_{2} = frac{7*4}{4} * a_{1} = frac{7*4 * 14}{4} = frac{2* 7^{2}*4 }{4}$

$a_{3} = frac{7*5}{6} * a_{2} = frac{2* 7^{3}*4*5 }{4*6}$

And here is the problem...
a4 equals:

$a_{4} = frac{7*6}{8} * a_{3} = frac{2* 7^{4}*4*5*6 }{4*6*8}$

We are multiplying numerator and denominator with 2*3 to get factorial 6! on numerator. I can't understand how the
$(1*2*3)^2$ got on denominator and how after that we transformed it to $3! * 4!$. Next steps are quite easy and the mathematical induction is fully understandable for me.

$a_{4} = frac{2* 7^{4}*6! }{ (1*2*3)^{2} * 2 *3 *4 } = frac{2* 7^{4}*6! }{ 3! 4! }$

Full answer of the exercise: JPG image

induction recursion

share|cite|improve this question

asked Mar 20 at 12:14

Jim.DJim.D

31

$endgroup$

add a comment | 

$begingroup$

I am struggling with that exercise... i don't understand a step in the answer. Maybe somebody Can explain in here or at least give me a hint what's going on there.

$a_{n} = frac{7*(n+2)}{2*n} *a_{n-1}$ ,$a_{1} = 14$

So it's...

$a_{2} = frac{7*4}{4} * a_{1} = frac{7*4 * 14}{4} = frac{2* 7^{2}*4 }{4}$

$a_{3} = frac{7*5}{6} * a_{2} = frac{2* 7^{3}*4*5 }{4*6}$

And here is the problem...
a4 equals:

$a_{4} = frac{7*6}{8} * a_{3} = frac{2* 7^{4}*4*5*6 }{4*6*8}$

We are multiplying numerator and denominator with 2*3 to get factorial 6! on numerator. I can't understand how the
$(1*2*3)^2$ got on denominator and how after that we transformed it to $3! * 4!$. Next steps are quite easy and the mathematical induction is fully understandable for me.

$a_{4} = frac{2* 7^{4}*6! }{ (1*2*3)^{2} * 2 *3 *4 } = frac{2* 7^{4}*6! }{ 3! 4! }$

Full answer of the exercise: JPG image

induction recursion

share|cite|improve this question

asked Mar 20 at 12:14

Jim.DJim.D

31

$endgroup$

share|cite|improve this question

asked Mar 20 at 12:14

Jim.DJim.D

31

share|cite|improve this question

asked Mar 20 at 12:14

Jim.DJim.D

31

asked Mar 20 at 12:14

Jim.DJim.D

31

asked Mar 20 at 12:14

Jim.DJim.D

31

3 Answers
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0

$begingroup$

Your doubt is valid:
$$begin{align}a_{n}& = frac{7*(n+2)}{2*n} *a_{n-1}, a_1=14\
a_2&=frac{7*4}{4}*a_1=frac{2*7^2*4}{4}=frac{2*7^2*4!}{4!}\
a_3&=frac{7*5}{6}*a_2=frac{2*7^3*4*5}{4*6}=frac{7^3*5!}{2*3!*3!}\
a_4&=frac{7*6}{8}*a_3=frac{7^4*5!*6}{2*3!*3!*8}=frac{7^4*6!}{2^2*3!*4!}\
a_5&=frac{7*7}{10}*a_4=frac{7^5*6!*7}{2^2*3!*4!*10}=frac{7^5*7!}{2^3*3!*5!}\
vdots\
a_n&=frac{7^n*(n+2)!}{2^{n-2}*3!*n!},nge 1.end{align}$$
Verify:
$$begin{align}a_1&=frac{7^1*(1+2)!}{2^{1-2}*3!*1!}=14\
a_2&=frac{7^2*(2+2)!}{2^{2-2}*3!*2!}=2*7^2\
a_3&=frac{7^3*(3+2)!}{2^{3-2}*3!*3!}=frac{7^3*5!}{2*3!*3!}end{align}$$

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answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank u a lot. Now is everything clear! Thanks once again :)
  $endgroup$
  – Jim.D
  Mar 21 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. If you think it answers your question, you can accept it. Good luck
  $endgroup$
  – farruhota
  Mar 21 at 14:24
  

  
  
  
  

add a comment | 

0

$begingroup$

I got $$a_n=frac{2}{3}frac{7^n}{2^n}(n+1)(n+2)$$

share|cite|improve this answer

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

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add a comment | 

0

$begingroup$

$$a_n=a_{n-r}cdotdfrac{7^r(n+2)(n+1)cdots(n-r+3)}{2^r n(n-1)cdots(n-r+1)}$$ $0le rle n-1$

For $rge2implies nge3$ $$a_n=a_{n-r}cdotleft(dfrac72right)^rdfrac{(n+2)(n+1)}{(n-r+2)(n-r+1)}$$

Set $r=n-1$

share|cite|improve this answer

edited Mar 20 at 12:41

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

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add a comment | 

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0

$begingroup$

Your doubt is valid:
$$begin{align}a_{n}& = frac{7*(n+2)}{2*n} *a_{n-1}, a_1=14\
a_2&=frac{7*4}{4}*a_1=frac{2*7^2*4}{4}=frac{2*7^2*4!}{4!}\
a_3&=frac{7*5}{6}*a_2=frac{2*7^3*4*5}{4*6}=frac{7^3*5!}{2*3!*3!}\
a_4&=frac{7*6}{8}*a_3=frac{7^4*5!*6}{2*3!*3!*8}=frac{7^4*6!}{2^2*3!*4!}\
a_5&=frac{7*7}{10}*a_4=frac{7^5*6!*7}{2^2*3!*4!*10}=frac{7^5*7!}{2^3*3!*5!}\
vdots\
a_n&=frac{7^n*(n+2)!}{2^{n-2}*3!*n!},nge 1.end{align}$$
Verify:
$$begin{align}a_1&=frac{7^1*(1+2)!}{2^{1-2}*3!*1!}=14\
a_2&=frac{7^2*(2+2)!}{2^{2-2}*3!*2!}=2*7^2\
a_3&=frac{7^3*(3+2)!}{2^{3-2}*3!*3!}=frac{7^3*5!}{2*3!*3!}end{align}$$

share|cite|improve this answer

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank u a lot. Now is everything clear! Thanks once again :)
  $endgroup$
  – Jim.D
  Mar 21 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. If you think it answers your question, you can accept it. Good luck
  $endgroup$
  – farruhota
  Mar 21 at 14:24
  

  
  
  
  

add a comment | 

0

$begingroup$

Your doubt is valid:
$$begin{align}a_{n}& = frac{7*(n+2)}{2*n} *a_{n-1}, a_1=14\
a_2&=frac{7*4}{4}*a_1=frac{2*7^2*4}{4}=frac{2*7^2*4!}{4!}\
a_3&=frac{7*5}{6}*a_2=frac{2*7^3*4*5}{4*6}=frac{7^3*5!}{2*3!*3!}\
a_4&=frac{7*6}{8}*a_3=frac{7^4*5!*6}{2*3!*3!*8}=frac{7^4*6!}{2^2*3!*4!}\
a_5&=frac{7*7}{10}*a_4=frac{7^5*6!*7}{2^2*3!*4!*10}=frac{7^5*7!}{2^3*3!*5!}\
vdots\
a_n&=frac{7^n*(n+2)!}{2^{n-2}*3!*n!},nge 1.end{align}$$
Verify:
$$begin{align}a_1&=frac{7^1*(1+2)!}{2^{1-2}*3!*1!}=14\
a_2&=frac{7^2*(2+2)!}{2^{2-2}*3!*2!}=2*7^2\
a_3&=frac{7^3*(3+2)!}{2^{3-2}*3!*3!}=frac{7^3*5!}{2*3!*3!}end{align}$$

share|cite|improve this answer

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank u a lot. Now is everything clear! Thanks once again :)
  $endgroup$
  – Jim.D
  Mar 21 at 14:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome. If you think it answers your question, you can accept it. Good luck
  $endgroup$
  – farruhota
  Mar 21 at 14:24
  

  
  
  
  

add a comment | 

$begingroup$

Your doubt is valid:
$$begin{align}a_{n}& = frac{7*(n+2)}{2*n} *a_{n-1}, a_1=14\
a_2&=frac{7*4}{4}*a_1=frac{2*7^2*4}{4}=frac{2*7^2*4!}{4!}\
a_3&=frac{7*5}{6}*a_2=frac{2*7^3*4*5}{4*6}=frac{7^3*5!}{2*3!*3!}\
a_4&=frac{7*6}{8}*a_3=frac{7^4*5!*6}{2*3!*3!*8}=frac{7^4*6!}{2^2*3!*4!}\
a_5&=frac{7*7}{10}*a_4=frac{7^5*6!*7}{2^2*3!*4!*10}=frac{7^5*7!}{2^3*3!*5!}\
vdots\
a_n&=frac{7^n*(n+2)!}{2^{n-2}*3!*n!},nge 1.end{align}$$
Verify:
$$begin{align}a_1&=frac{7^1*(1+2)!}{2^{1-2}*3!*1!}=14\
a_2&=frac{7^2*(2+2)!}{2^{2-2}*3!*2!}=2*7^2\
a_3&=frac{7^3*(3+2)!}{2^{3-2}*3!*3!}=frac{7^3*5!}{2*3!*3!}end{align}$$

share|cite|improve this answer

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

$endgroup$

share|cite|improve this answer

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

answered Mar 20 at 13:20

farruhotafarruhota

21.8k2842

0

$begingroup$

I got $$a_n=frac{2}{3}frac{7^n}{2^n}(n+1)(n+2)$$

share|cite|improve this answer

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

add a comment | 

0

$begingroup$

I got $$a_n=frac{2}{3}frac{7^n}{2^n}(n+1)(n+2)$$

share|cite|improve this answer

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

add a comment | 

$begingroup$

I got $$a_n=frac{2}{3}frac{7^n}{2^n}(n+1)(n+2)$$

share|cite|improve this answer

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

share|cite|improve this answer

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

answered Mar 20 at 12:33

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

0

$begingroup$

$$a_n=a_{n-r}cdotdfrac{7^r(n+2)(n+1)cdots(n-r+3)}{2^r n(n-1)cdots(n-r+1)}$$ $0le rle n-1$

For $rge2implies nge3$ $$a_n=a_{n-r}cdotleft(dfrac72right)^rdfrac{(n+2)(n+1)}{(n-r+2)(n-r+1)}$$

Set $r=n-1$

share|cite|improve this answer

edited Mar 20 at 12:41

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

0

$begingroup$

$$a_n=a_{n-r}cdotdfrac{7^r(n+2)(n+1)cdots(n-r+3)}{2^r n(n-1)cdots(n-r+1)}$$ $0le rle n-1$

For $rge2implies nge3$ $$a_n=a_{n-r}cdotleft(dfrac72right)^rdfrac{(n+2)(n+1)}{(n-r+2)(n-r+1)}$$

Set $r=n-1$

share|cite|improve this answer

edited Mar 20 at 12:41

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

$begingroup$

$$a_n=a_{n-r}cdotdfrac{7^r(n+2)(n+1)cdots(n-r+3)}{2^r n(n-1)cdots(n-r+1)}$$ $0le rle n-1$

For $rge2implies nge3$ $$a_n=a_{n-r}cdotleft(dfrac72right)^rdfrac{(n+2)(n+1)}{(n-r+2)(n-r+1)}$$

Set $r=n-1$

share|cite|improve this answer

edited Mar 20 at 12:41

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

share|cite|improve this answer

edited Mar 20 at 12:41

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 20 at 12:32

lab bhattacharjeelab bhattacharjee

228k15158279