Question

I posted this one a few days ago, but no oe answered it. Please try.
Consider the table
1 = 0 + 1
1 +3 +4 = 1 + 8
5 + 6 + 7+8+9 = 8 + 27
10+11+12+13+14+15+16 = 27 +64
each line has the next odd number of consecutive integers, and each line equals the sum of the last perfect cube used and the next consecutive cube. but I can't figure out how to write this in suitable mathematical form, let alone prove it.

Answer 1

this looks like an identity of some sort
but i notice you have omitted the number 2

Answer 2

oh the second line is 2 + 3+ 4, right?

Answer 3

i think 2 was there in the second line becoz
2 + 3 + 4 = 1+8

Answer 4

yes, sorry for the typo

Answer 5

they are all arithmetic series with d = 1 and different first term

Answer 6

but where do we go from there?

Answer 7

but how would you write it in mathematical form so that you could proove what the next line would be?

Answer 8

hmm - good question!!

Answer 9

This is problem #2 in Chapter one in George Polya"s Introduction and Analogy in Mathematics,Vol 1.
I think it's supposed to be solved with induction. Can we do induction on a list of arithmetic series?

Answer 10

title of book is Induction and Analogy

Answer 11

here's some bits:-

sum of first k digits k(k+1)/2 and sum of first k cubes is

/k(k+1)/2)^2

Answer 12

i expect we can

Answer 13

I like where we're going. But how do we indicate that the next line starts at the next consecutive number each time, and that the cubes are also consecutive?

Answer 14

i don't think you expression for cubes works
0 + 1)/2 =0
1^3 + 2^3)/2 =3

Answer 15

i'm not ducking out of this tad but I have to play taxi driver to my grandaughter i'll be back later

Answer 16

= 4.5 not 3

Answer 17

thanks jimmy

Answer 18

(k(k+1)/2)^2
1(1+1)/2 squared = 1
2(2+1)/2 squared = 9

Answer 19

but the first cube is 0

Answer 20

Number theory doesn't consider 0

Answer 21

this problem does

Answer 22

Just start your induction at 1...

Answer 23

okay; if I do that, how do I indicate that each line has different numbers, and a different number of numbers

Answer 24

The last number in a line n on the lhs is n^2 and the rest of the numbers are consecutive down to (n-1)^2 +1

The rhs for a line n is 1 + 2^3 + 3^3 +....n^3

Answer 25

that's cool. I think i can get some where. Be back in a while