Al P. answered • 09/29/19

Online Mathematics tutor

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

Assume

n

∑ (2k-1)^{2} = m(4m^{2} - 1) /3 is true for n=m. This is the hypothesis.

k

Now let n=m+1:

m+1 m

∑ (2k-1)^{2} = ∑ (2k-1)^{2} + (2(m+1)-1)^{2}

k=1 k=1

where the RHS has been broken into two terms such that we can apply the hypothesis to the first term on the RHS:

m+1

∑ (2k-1)^{2} = m(4m^{2}-1) /3 + (2(m+1)-1)^{2}

k=1

The RHS expands to **(4m**^{3}**+12m**^{2}**+11m+3)/3**

Now, one way to go is to check if this is the same as m+1 plugged into n(4n^{2}-1)/3

(m+1)(4(m+1)^{2} - 1)/3 =

(m+1)(4m^{2}+8m+3)/3 =

( 4m^{3}+8m^{2}+3m+4m^{2}+8m+3 )/3 =

** (4m**^{3}**+12m**^{2}**+11m+3)/3**

So, yes, the two expressions are the same and the proof is complete.

Another way to go is to factor (4m^{3}+12m^{2}+11m+3)/3:

(4m^{3}+12m^{2}+11m+3)/3 =

(m+1)(2m+1)(2m+3) / 3 =

(m+1)(4m^{2}+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).