1. Proof By Induction

For my Linear Algebra class we have to do Proof By Induction problems. I got #1 and #2 on the first page and I got the whole second page. But what do you do on #3 and #4 on the first page? Images attached. Thanks you for your time - I appreciate it.

2. Problem 1

$\displaystyle \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Base Case

$\displaystyle \sum_{i=1}^{1} i^2 = 1^2 = 1 , \frac{1(1+1)(2*1+1)}{6} = \frac{6}{6} = 1$

Assume k $\displaystyle \in$ S

$\displaystyle \sum_{i=1}^{k} i^2 = \frac{n(n+1)(2n+1)}{6}$

Prove k+1 $\displaystyle \in$ S

$\displaystyle \sum_{i=1}^{k+1} i^2 = \frac{(k+1)(k+2)(2k+3)}{6} = \frac{(k+1)(2k^2+7K+6)}{6} = \frac{2k^3+9k^2+13k+6}{6}$
$\displaystyle \sum_{i=1}^{k+1} i^2 = \sum_{i=1}^{k} i^2 + (k+1)^2$
$\displaystyle \sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + k^2+2k+1$
$\displaystyle = \frac{(k^2+k)(2k+1)}{6} + \frac{6k^2+12k+6}{6} = \frac{2k^3+3k^2+k}{6} + \frac{6k^2+12k+6}{6}$
$\displaystyle = \frac{2k^3+9k^2+13k+6}{6}$

3. For the 3rd one,

$\displaystyle 1!<2^1$
$\displaystyle 2!<2^2$
$\displaystyle 3!<2^2$
$\displaystyle 4!>2^4$

If $\displaystyle n!>2^n,\$ then $\displaystyle (n+1)!>2^{n+1}$

$\displaystyle (n+1)!=(n+1)n(n-1)(n-2)(n-3)... =(n+1)n!$

$\displaystyle (n+1)n!>2^{n+1}$ ?

$\displaystyle (n+1)n!>(2)2^n$ ?

As n+1>2 for n>2

$\displaystyle (n+1)!>2^{n+1}$ if $\displaystyle n!>2^n$

4. For 4,

$\displaystyle \sum_{i=0}^na^i=\frac{1-a^{n+1}}{1-a},\ a\ne1$

$\displaystyle n=0,\ a^0=\frac{1-a}{1-a}$

$\displaystyle n=1,\ a^0+a^1=\frac{1-a^2}{1-a}=\frac{(1-a)(1+a)}{1-a}=1+a$

If $\displaystyle \sum_{i=0}^na^i=\frac{1-a^{n+1}}{1-a}$ then $\displaystyle \sum_{i=0}^{n+1}a^i=\frac{1-a^{n+2}}{1-a}$

Proof

$\displaystyle \frac{1-a^{n+1}}{1-a}+a^{n+1}=\frac{1-a^{n+1}+(1-a)a^{n+1}}{1-a}$

$\displaystyle =\frac{1-a^{n+1}+a^{n+1}-a^{n+2}}{1-a}=\frac{1-a^{n+2}}{1-a}$

5. Finally,

1=1
1+3=4
1+3+5=9
1+3+5+7=16

$\displaystyle \sum_{i=1}^n2n-1=n^2$

Hence $\displaystyle \sum_{i=1}^n2(n+1)-1$ should $\displaystyle =(n+1)^2$

$\displaystyle \sum_{i=1}^n2n+1=n^2+2n+1$ ?

Proof

$\displaystyle \sum_{i=1}^n2n-1+[2(n+1)-1]=n^2+[2(n+1)-1]=n^2+2n+1$