Diferencia entre revisiones de «Práctica 1: Inducción (Algoritmos III)»

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Línea 1: Línea 1:
==Ejercicio 01.01:==
==Ejercicio 01.01:==
<br>a)
<br>a)
P(n) = Σ<sub>i=1..n</sub> i = n(n+1)/2
P(n) = Σ{i=1..n} i = n(n+1)/2
* CB: n = 1
* CB: n = 1


Σ<sub>i=1..1</sub> i = 1
Σ{i=1..1} i = 1


1(1+1)/2 = 1 OK
1(1+1)/2 = 1 OK


* PI: P(n)=>P(n+1)
* PI: P(n)=>P(n+1)
Σ<sub>i=1..n+1</sub> i = Σ<sub>i=1..n</sub> i + n+1 = (HI) n(n+1)/2 + 2(n+1)/2 = (n+2)(n+1)/2 OK
Σ{i=1..n+1} i = Σ{i=1..n} i + n+1 = (HI) n(n+1)/2 + 2(n+1)/2 = (n+2)(n+1)/2 OK


<br>b)
<br>b)
P(n) = Σ<sub>i=0..n</sub> (2*i+1) = (n+1)<sup>2</sup>
P(n) = Σ{i=0..n} (2*i+1) = (n+1)^2
* CB: n = 0
* CB: n = 0


Σ<sub>i=0..0</sub> (2*i+1) = 1
Σ{i=0..0} (2*i+1) = 1


(0+1)<sup>2</sup> = 1 OK
(0+1)^2 = 1 OK


* PI: P(n)=>P(n+1)
* PI: P(n)=>P(n+1)
Σ<sub>i=0..n+1</sub> (2*i+1) = Σ<sub>i=0..n</sub> 2*i+1 + 2*(n+1)+1 = (HI) (n+1)<sup>2</sup> + 2n + 3 = n<sup>2</sup> + 2*n + 1 + 2n + 3 = n<sup>2</sup> + 4*n + 4 = (n+2)<sup>2</sup> OK
Σ{i=0..n+1} (2*i+1) = Σ{i=0..n} 2*i+1 + 2*(n+1)+1 = (HI) (n+1)^2 + 2n + 3 = n^2 + 2*n + 1 + 2n + 3 = n^2 + 4*n + 4 = (n+2)^2 OK
 
<br>c)
<br>c)
P(n) = Σ{i=1..n} i^2 = n(n+1)(2n+1)/6
* CB: n = 1
Σ{i=1..1} i^2 = 1
1(1+1)(2*1+1)/6 = 1 OK
* PI: P(n)=>P(n+1)
Σ{i=0..n+1} i^2 = Σ{i=0..n} i^2 + (n+1)^2 = (HI) n(n+1)(2n+1)/6 + (n+1)^2 = [ n(n+1)(2n+1)/6 + 6*(n+1)^2 ]/6 = (n+1)[ n(2n+1)/6 + 6*(n+1) ]/6 = (n+1)[ 2*n^2 + n + 6*n + 6 ]/6 = (n+1)(n+2)(2n+3)/6 OK
<br>d)
<br>d)
<br>e)
<br>e)
<br>f)
<br>f)



Revisión del 21:32 11 nov 2006

Ejercicio 01.01:


a) P(n) = Σ{i=1..n} i = n(n+1)/2

  • CB: n = 1

Σ{i=1..1} i = 1

1(1+1)/2 = 1 OK

  • PI: P(n)=>P(n+1)

Σ{i=1..n+1} i = Σ{i=1..n} i + n+1 = (HI) n(n+1)/2 + 2(n+1)/2 = (n+2)(n+1)/2 OK


b) P(n) = Σ{i=0..n} (2*i+1) = (n+1)^2

  • CB: n = 0

Σ{i=0..0} (2*i+1) = 1

(0+1)^2 = 1 OK

  • PI: P(n)=>P(n+1)

Σ{i=0..n+1} (2*i+1) = Σ{i=0..n} 2*i+1 + 2*(n+1)+1 = (HI) (n+1)^2 + 2n + 3 = n^2 + 2*n + 1 + 2n + 3 = n^2 + 4*n + 4 = (n+2)^2 OK


c) P(n) = Σ{i=1..n} i^2 = n(n+1)(2n+1)/6

  • CB: n = 1

Σ{i=1..1} i^2 = 1

1(1+1)(2*1+1)/6 = 1 OK

  • PI: P(n)=>P(n+1)

Σ{i=0..n+1} i^2 = Σ{i=0..n} i^2 + (n+1)^2 = (HI) n(n+1)(2n+1)/6 + (n+1)^2 = [ n(n+1)(2n+1)/6 + 6*(n+1)^2 ]/6 = (n+1)[ n(2n+1)/6 + 6*(n+1) ]/6 = (n+1)[ 2*n^2 + n + 6*n + 6 ]/6 = (n+1)(n+2)(2n+3)/6 OK


d)


e)


f)

Ejercicio 01.02:

HI = Σi=0..n 2i = 2n+1-1

  • CB: n = 0

Σi=0..0 2i = 1

20+1-1 = 2-1 = 1 OK

  • PI: P(n)=>P(n+1)

Σi=0..n+1 2i = Σi=0..n 2i + 2n+1 = (HI) 2n+1-1 + 2n+1 = 2 * 2n+1-1 = 2n+2-1 OK

Ejercicio 01.03:

Ejercicio 01.04:

Ejercicio 01.05:

Ejercicio 01.06:

Ejercicio 01.07: