My induction does not prove my conjecture. Is my conjecture wrong or is my induction insufficient/wrong?

Question

I am asked to consider the following sequence: $$1=0+1\\2+3+4=1+8\\5+6+7+8+9=8+27\\10+11+12+13+14+15+16=27+64$$ The questions are:

a) What is the next equality in this sequence?

b) What conjecture is suggested by these equalities?

c) Prove the conjecture in (b) by induction.

Here are my answers:

a) $17+18+19+20+21+22+23+24+25=27+64$

b) $\forall i \in \mathbb{Z}^+, \sum_{k=(i-1)^2+1}^{i^2}k=(i-1)(i-1)^2+(i)^3$

c) $$\text{Proof: We proceed by induction.} \\\text{Base case: } \sum_{k=(1-1)^2+1}^{1^2}k=1, (1-1)(1-1)^2+(1)^3=0+1=1, \\\text{Thus: } LHS=RHS \\\text{We assume for some integer z, } \sum_{k=(z-1)^2+1}^{z^2}k=(z-1)(z-1)^2+(z)^3 \\\text{We need to show, } \sum_{k=(z-1)^2+1}^{(z+1)^2}k=(z)^3+(z+1)^3 \\\text{Hence, } \sum_{k=(z-1)^2+1}^{z^2}k + (z+1)^2 = \sum_{k=(z-1)^2+1}^{(z+1)^2}k=(z-1)(z-1)^2+(z)^3+(z+1)^2$$

But from there I can't get:

$$(z-1)(z-1)^2+(z)^3+(z+1)^2=(z)^3+(z+1)^3$$

Maybe I am misunderstanding something here that should be obvious but I am still learning to prove by induction. Using strong induction has gone through my mind but at this point I don't feel confident enough in my understanding of strong induction to use it here. (I am self-learning this.)

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Update: This is how I proved it by using the index of the summation.

We notice in what we want to show, that if we change the indexing to zero, then we can write the summation as follows:

$$ \text{Let } p\ge 0 \text{.} \\ \sum_{k=(p)^2+1}^{(p+1)^2}k=((p+1)-1)((p+1)-1)^2+(p+1)^3 = (p)^3+(p+1)^3 \blacksquare$$

Answer 1

Your approach is correct but you have some minor mistakes in your approach.

For example $$17+18+19+20+21+22+23+24+25=27+64$$ should have been

$$17+18+19+20+21+22+23+24+25=64+125$$

and $$\text{We need to show, } \sum_{k=(z-1)^2+1}^{(z+1)^2}k=(z)^3+(z+1)^3$$

Should have been $$\text{We need to show, } \sum_{k=z^2+1}^{(z+1)^2}k=(z)^3+(z+1)^3$$

Answer 2

In this case, I think a direct solution is easier than induction (though surely induction should work)

Index the formulas starting with $n=0$. The $n^{th}$ left hand sum can be written as: $$m(n)-n,m(n)-n+1,\cdots,m(n)-1, m(n),m(n)+1,\cdots, m(n)+n$$

for $m(n)=n^2+n+1$.

It follows that the sum of the terms in the $n^{th}$ sum is $$(2n+1)m(n)=(2n+1)(n^2+n+1)=2n^3+3n^2+3n+1$$

On the other hand the $n^{th}$ right hand is $$(n+1)^3+(n)^3=2n^3+3n^2+3n+1$$ and we are done.