Properties of integers:math induction

AI Thread Summary
The discussion revolves around deriving a general formula from four given equations involving sums of integers and their relationships to cubes. Participants identify that the left side of the equations represents a sum of consecutive integers, while the right side relates to cubic numbers. A conjecture is proposed that the sum can be expressed as n^2 - 2(n - 1) + ... + n^2 = (n-1)^3 + n^3, which is verified for specific values of n. The need for a proof by induction is acknowledged, emphasizing that the conjecture holds true for the provided equations. The conversation highlights the complexity of formulating a general proof beyond initial observations.
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Homework Statement


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


Homework Equations





The Attempt at a Solution



I know the left side has n + (n+1) numbers,
and the right side has the last number occurring the in next equation,
i also see that the numbers on the right side come from an k^3 number.
but how would i write this in a general formula which would be sufficient to
satisfy all the equations? because to me it seems like a recursive definition but
i don't quite clearly understand how i would write out an answer for this...
and also what exactly am i supposed to write down to solve this??
 
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Notice also that the last number in the sum on the left is always a square, say, n2 and the last number on the right is n3. Use that as your base.
It looks to me like the first number on the left is (n-1)2+ 1 so the sum on the left is from (n-1)2+ 1 to n2.
 
so on the left side... could i use sigma notation to verify the range and the addition of the numbers?? and on the right side ... i can see that it's just like
1^3,
1^3 + 2^3 ,
2^3 + 3^3
3^3 + 4^3
how would i show this in terms of just using n ?
n + (n)^3 ?

the problem with this is that it will only apply to one specific instant of the equation,
or am i wrong??
 
On both left and right side you are going to have to specify what "n" is!
 
I guess the conjecture is
n^2 - 2 (n - 1) +...+ n^2 = (n-1)^3 + n^3
and it holds by trivial processing of the two sides
 
bmannaa said:
I guess the conjecture is
n^2 - 2 (n - 1) +...+ n^2 = (n-1)^3 + n^3
Does your guess work for the given equations?
bmannaa said:
and it holds by trivial processing of the two sides
There's a lot more to this than "trivial processing." Note the title of the thread.
 
Mark44 said:
Does your guess work for the given equations?
Yes it does
in the formula n^2 - 2 (n - 1) +...+ n^2 = (n-1)^3 + n^3
1) 1=1
put n = 1
2) 2 + 3 + 4 = 1 + 8
put n = 2
3) 5 + 6 + 7 + 8 + 9 = 8 + 27
put n = 3
4) 10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64
put n = 4
Mark44 said:
There's a lot more to this than "trivial processing." Note the title of the thread.
Yep! I didn't get to prove it by induction.
 
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