Inductive proofs follow this general recipe:
- You first show it to be true for some n (usually n=1).
- Then you assume it is true for n = m. This is the hypothesis, and it is often used in the proof.
- Then you must show that assuming it to be tru for n = m, leads to it being true for n = m+1
For the posted problem....
n = 1: summation is 1, right hand side is 1(4-1)/3 = 3/3 = 1 so it is true for n=1
∑ (2k-1)2 = m(4m2 - 1) /3 is true for n=m. This is the hypothesis.
Now let n=m+1:
∑ (2k-1)2 = ∑ (2k-1)2 + (2(m+1)-1)2
where the RHS has been broken into two terms such that we can apply the hypothesis to the first term on the RHS:
∑ (2k-1)2 = m(4m2-1) /3 + (2(m+1)-1)2
The RHS expands to (4m3+12m2+11m+3)/3
Now, one way to go is to check if this is the same as m+1 plugged into n(4n2-1)/3
(m+1)(4(m+1)2 - 1)/3 =
( 4m3+8m2+3m+4m2+8m+3 )/3 =
So, yes, the two expressions are the same and the proof is complete.
Another way to go is to factor (4m3+12m2+11m+3)/3:
(m+1)(2m+1)(2m+3) / 3 =
(m+1)(4m2+8m+3) / 3 =
Noting that the 2nd factor is 4(m+1)2-1:
(m+1)(4(m+1)2- 1) / 3 (done, recall n=m+1 so you can replace "m+1" with "n" and end up with the hypothesis expression).