#import "@youwen/zen:0.1.0": *

#show: zen.with(
  title: "Homework 5",
  author: "Youwen Wu",
)

#set heading(numbering: none)
#show heading.where(level: 2): it => [#it.body.]
#show heading.where(level: 3): it => [#it.body.]

#set par(first-line-indent: 0pt, spacing: 1em)


Problems:

2.4: \#4bcde, 5ajq, 6ce, 7a, 9, 10, 12abc

2.5: \#1abc, 3, 10

#outline()

= 2.4

== 4

=== b

$3 + 11 + 19 + dots.c + (8n - 5) = 4n^2 - n$

First we show the base case for $n = 1$.

$ 3 = 4 (1^2) - 1 = 3 $

Now we proceed by induction. Assume $sum _(k=1) ^n (8k-5) = 4(n^2) - n$.

$
  sum_(k=1)^(n+1) (8k-5) &= 4((n+1)^2) - (n+1) \
  (sum_(k=1)^(n) (8k-5)) + (8(n+1) - 5) &= 4(n^2 + 2n + 1) - n-1 \
  (sum_(k=1)^(n) (8k-5)) + 8n + 3 &= 4n^2 - n + 8n + 3 \
  (sum_(k=1)^(n) (8k-5)) &= 4n^2 - n \
$

and we know that this is true by our original assumption. So by the PMI, this
is true $forall n in NN$.

=== c

$sum_(i=1)^n 2^i = 2^(n+1) - 2$

For the base case:

$ 2^1 = 2(1 + 1) - 2 \ 2 = 2 $

Assume that

$
  sum_(i=1)^n 2^i = 2^(n+1) - 2
$

Proceeding by induction,

$
  sum_(i=1)^(n+1) 2^i = 2^(n+1+1) - 2 \
  (sum_(i=1)^(n) 2^i) + 2^(n+1) = 2^(n+2) - 2 \
  (sum_(i=1)^(n) 2^i) = 2^(n+2) - 2^(n+1) - 2 \
  (sum_(i=1)^(n) 2^i) = 2^(n+1) dot (2 - 1) - 2 \
  (sum_(i=1)^(n) 2^i) = 2^(n+1) - 2 \
$

So by the PMI, it holds for all $n in NN$.

=== d

$1 dot 1! + 2 dot 2! + 3 dot 3! + dots.c + n dot n! = (n + 1)! - 1$

For the base case: $1 dot 1! = (1 + 1)! - 1 \ 1 = 1$

Assuming the inductive hypothesis,

$
  sum^n_(k=1) k dot k! = (n + 1)! - 1 \
$

Now we proceed by induction

$
  sum^(n+1)_(k=1) k dot k! = (n + 2)! - 1 \
  sum^(n)_(k=1) k dot k! + (n + 1) dot (n+1)! = (n + 2) dot (n + 1)! - 1 \
  sum^(n)_(k=1) k dot k! = (n + 2) dot (n + 1)! - (n + 1) dot (n+1)! - 1 \
  sum^(n)_(k=1) k dot k! = (n + 1)! dot (n+2 - (n + 1) - 1 \
  sum^(n)_(k=1) k dot k! = (n + 1)! - 1 \
$

So by the PMI, this is true for all $n in NN$