Proof by mathematical induction

Answer: 3k2+6k+3=3(k+1)2

1. At n=1 both sides equal 3.
2. Add the next term, 3(2k+1), to the assumed sum.
3. Factor 3(k2+2k+1)=3(k+1)2.

Answer: 4k+1−1=3(4m+1)

1. The base case gives 4−1=3.
2. Substitute 4k=3m+1 into 4×4k−1.
3. This gives 12m+3=3(4m+1), proving divisibility.

Answer: 5k2+10k+5=5(k+1)2

1. At n=1 both sides equal 5.
2. Add the next term, 5(2k+1), to the assumed sum.
3. Factor 5(k2+2k+1)=5(k+1)2.