Hey can you guys verify if I did these proofs by induction practice problems correct? Just going to type out simplified versions:

Part 1:

(3n-4)=(n/2)(3n-5)

(3(1)-4)=((1)/2)(3(1)-5)

3-4=1/2(3-5)

-1=1/2(-2)

-1=-1

Part 2:

(3n-4)=(n/2)(3n-5)

3k-4=(k/2)(3k-5)

(3k-4)+(3(k+1)-4)=((k+1)/2)(3(k+1)-5)

(k/2)(3k-5)+(3(k+1)-4)=((k+1)/2)(3(k+1)-5)

3k^2/2-5k/2+3k-1=((k+1)/2)(3k-2)

3k^2/2-5k/2+3k-1=(3k(k+1)/2)-(2(k+1)/2)

3k^2/2-5k/2+3k-1=3k^2/2+3k/2-k-1

3k^2/2-5k/2+6k/2-1=3k^2/2+3k/2-k-1

3k^2/2+k/2-1=3k^2/2+3k/2-k-1

3k^2/2+k/2-1=3k^2+3k/2-2k/2-1

3k^2/2+k/2-1=3k^2+k/2-1
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Part 1:

(2n-1)=n^2

2(1)-1=(1)^2

2-1=1

1=1

Part 2:

(2n-1)=n^2

(2k-1)=k^2

(2k-1)+(2(k+1)-1)=(k+1)^2

(2k-1)+(2(k+1)-1)=(k+1)^2

k^2+2k+1=k^2+2k+1
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Part 1:

(4n-1)=n(2n+1)

(4(1)-1)=(1)(2(1)+1)

4-1=2+1

3=3

Part 2:

(4n-1)=n(2n+1)

(4k-1)=k(2k+1)

(4k-1)+(4(k+1)-1)=(k+1)(2(k+1)+1)

k(2k+1)+(4(k+1)-1)=(k+1)(2k+3)

k(2k+1)+(4(k+1)-1)=2k^2+5k+3

2k^2+k+4k+3=2k^2+5k+3

2k^2+5k+3=2k^2+5k+3