# Mathematical Induction

• Feb 22nd 2008, 12:31 AM
BACONATOR
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(Talking)
Thanks!
• Feb 22nd 2008, 01:25 AM
afeasfaerw23231233
here's my working
• Feb 22nd 2008, 01:40 AM
CaptainBlack
Quote:

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(Talking)
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
• Feb 22nd 2008, 08:45 AM
Soroban
Hello, BACONATOR!

Find the general statement is the biggest task . . .

Quote:

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

• Jul 12th 2013, 10:18 AM
Soroban
Re: Mathematical Induction
Hello, BACONATOR!

Quote:

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

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

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

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

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

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

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

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

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

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

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

. . . . . . . . . $\displaystyle S \;=\;(k-1)^3 + k^3$
• Jul 12th 2013, 10:44 AM
topsquark
Re: Mathematical Induction
I'm curious. You replied to your own post from 5 years ago with an almost identical one?

-Dan