1. ## Mathematical Induction

Hello!
I've got this problem.

18. Consider the following four equations:

1) 1 = 1
2) 2+3+4 = 1+8
3) 5+6+7+8+9 = 8+27
4) 10+11+12+13+14+15+16 = 27+64
Conjecture the general formula suggested by these four equations, and prove your conjecture.

Any help would be appreciated
Thanks!

2. here's my working

3. Originally Posted by BACONATOR
Hello!
I've got this problem.

18. Consider the following four equations:

1) 1 = 1
2) 2+3+4 = 1+8
3) 5+6+7+8+9 = 8+27
4) 10+11+12+13+14+15+16 = 27+64
Conjecture the general formula suggested by these four equations, and prove your conjecture.

Any help would be appreciated
Thanks!
Conjecture: $\displaystyle \forall n \in \mathbb{N}$

$\displaystyle \sum_{r=(n-1)^2+1}^{n^2} r = (n-1)^3+n^3$

and you should not need induction to prove this as the left hand side is an
arithmetic progression.

RonL

4. Hello, BACONATOR!

Find the general statement is the biggest task . . .

18. Consider the following four equations:

$\displaystyle \begin{array}{cccccc} 1) & 1 & = & 1 & = & 0^3+1^3\\ 2) & 2+3+4 &= & 1+8 & = & 1^3+2^3\\ 3) & 5+6+7+8+9 & = & 8+27 &=& 2^3+3^3\\ 4) & 10+11+12+13+14+15+16 & = & 27+64 & = & 3^3+4^3 \end{array}$

Conjecture the general formula suggested by these four equations,
Here's the pattern that I saw . . .

Equation 2: sum of 3 consecutive integers up to 2², equals 1³ + 2³

Equation 3: sum of 5 consecutive integers up to 3², equals 2³ + 3³

Equation 4: sum of 7 consecutive integers up to 4², equals 3³ + 4³

My conjecture:

Equation $\displaystyle n$: sum of $\displaystyle 2n-1$ consecutive integers up to $\displaystyle n^2$, equals $\displaystyle (n-1)^3 + n^3$

After a bit of algebra, I found that equation $\displaystyle n$
. . has a sum that begins with $\displaystyle n^2-2n+2$

The sum is an arithmetic series with first term, $\displaystyle a \:= \:n^2-2n+2$,
. . common difference, $\displaystyle d = 1$, and $\displaystyle 2n-1$ terms.

This sum is: .$\displaystyle S \;=\;\frac{2n-1}{2}\left[2(n^2-2n+2) + 2n(1)\right]$

. . which simplifies to: .$\displaystyle 2n^3 - 3n^2 + 3n - 1 \;=\;(n-1)^3 + n^3$

5. ## Re: Mathematical Induction

Hello, BACONATOR!

$\displaystyle \text{18. Consider the following four equations:}$

$\displaystyle \begin{array}{cccccc}(1) & 1 &=& 0^3 + 1^3 \\(2) & 2+3+4 &=& 1^3+2^3 \\ (3) & 5+6+7+8+9 &=& 2^3+3^3 \\ (4) & 10+11+12+13+14+15+16 &=& 3^3+4^3 \end{array}$

$\displaystyle \text{Conjecture the general formula suggested by these four equations,}$
. . $\displaystyle \text{and prove your conjecture.}$

The left side is an arithmetic series.
. . Its first term is: $\displaystyle a = k^2 - 2k+2$
. . The common difference is: $\displaystyle d = 1$,
. . The number of terms is: $\displaystyle n \,=\,2k-1$

The sum is: .$\displaystyle S \;=\;\frac{n}{2}\big[2a + (n-1)d\big]$

. . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2(k^2-2k+2) + (2k-2)1\big]$

. . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2k^2 - 4k+ 4 + 2k - 2\big]$

. . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2k^2 - 2k + 2\big]$

. . . . . . . . . $\displaystyle S \;=\;(2k-1)(k^2 - k + 1)$

. . . . . . . . . $\displaystyle S \;=\;2k^3 - 2k^2 + 2k - k^2 + k - 1$

. . . . . . . . . $\displaystyle S \;=\; 2k^3 - 3k^2 + 3k - 1$

. . . . . . . . . $\displaystyle S \;=\;(k^3 - 3k^2 + 3k - 1) + k^3$

. . . . . . . . . $\displaystyle S \;=\;(k-1)^3 + k^3$

6. ## Re: Mathematical Induction

I'm curious. You replied to your own post from 5 years ago with an almost identical one?

-Dan