HW3	Kishan Patel	9/19/21

1)
a)
sum(2k-1)
f(1)=2*1-1=1
f(2)=2*1-1=3+1=4
1=1^2
4=2^2
n => n+1
sum(2n-1)+2(n+1)-1
n^2+2n+1
(n+1)^2

b)
2k-1 is kth odd number

first odd number 1		
			
		5		
		5	
	3	5		
	3	5	
1	3	5	

	

5	5	5
3	3	5
1	3	5

n^2

c)
n=1  :=	 2*1-1 = 1^2		True
n=2  :=  2*2-1+1 = 2^2		True
n=3  :=	 3*2-1+4+1 = 3^2	True

2)
a)
1=1/2(1)(1+1) true
sum(k+n+1)
1/2n(n+1)+(n+1)2/2
1/2(n+1)(n+2)
thus this can be simplified to show that n+1 via indution is the same as n

b)
assume n=6
1 + 2 + 3 + 4 + 5 + 6 = 21
6 + 5 + 4 + 3 + 2 + 1

6(6+1)/2 = 21

c)
n = 1 := 1 = 1(2)/2 True
n = 2 := 2+1 = 2(2+1)/2 True
n = 3 := 3+2+1 = 3(3+1)/2 True
k=0 := 0	=0
k=1 := 1^2 	=1
k=2 := 2^2+1	=5
k=3 := 3^2+4+1	=14
k=4 := 4^2+9+4+1=30

n*(n+1)(an+b)
k=2 := 	2/5=2a+b
k=3 := 	3/12=3a+b
	-3/20=a
	2/5=-3/10+b
b=7/10

n*(n+1)(-3/20n+3/10)

4)
k=1	:=	1^3		=1
k=2	:=	2^3+1		=9
k=3	:=	3^3+8+1		=36
k=4	:=	4^3+27+8+1	=100
n=1	:=	(1(1+1)/2)^2	=1
n=2	:=	(2(2+1)/2)^2	=9
n=3	:=	(3(3+1)/2)^2	=26
n=4	:=	(4(4+1)/2)^2	=100