Maths question???

Question

some square numbers can be divided up into eight TRIANGLE numbers.
Some square numbers can be divided up into eight triangly numbers PLUS ON

which square numbers can?

(square numbers are 1, 4, 9, 16 , 25 , 36 , 49...)

I've been stuck on this question for ages so any help would be apreciated!

Which numbers can be represented as trapeziods with two addends?
three addends?
'n' addends?'

Can powers of 2 be trapeziodal numbers?
justify your answer.

there is a connection between trapeziodal numbers and the area of a trapezium


thanks a lot!

Answer 1

In answer to your first question, triangular numbers
can be represented as n(n + 1)/2 for n = 1, 2, 3, ...
Multiplying this by 8 and adding 1 gives:
8n(n + 1)/2 + 1
= 4n(n + 1) + 1
= 4n^2 + 4n + 1
= (2n + 1)^2
Now 2n + 1 represents all the odd numbers >= 3.
Thus, if you square all the odd numbers,
each is equal to 8 times a triangular number + 1.
So we have: (2n + 1)^2 = 8 * [n(n + 1)/2] + 1

n = 1 implies 3^2 = 8 * 1 + 1
n = 2 implies 5^2 = 8 * 3 + 1
n = 3 implies 7^2 = 8 * 6 + 1
n = 4 implies 9^2 = 8 * 10 + 1

Or we can give the general formula as :
(2n + 1)^2 = 8 * T(n) + 1, for n = 1, 2, 3, ...
where T(n) is the n(th) triangular number.

EDIT 1: On another question you posed -
Can powers of 2 be trapezoidal numbers?

Trapezoidal numbers are numbers that are
additions of consecutive numbers.
Let the consecutive numbers addition be:
(n) + (n + 1) + (n + 2) + ... + (n + r) = 2^x
There are (r + 1) terms here. Adding gives:
(r + 1)n + r(r + 1)/2 = 2^x
or, (r + 1)(n + r/2) = 2^x.
Multiplying through by 2 gives:
(r + 1)(2n + r) = 2^(x + 1), which is still a power of 2.
If r is odd, then (r + 1) is even, but (2n + r) is odd.
If r is even, then (r + 1) is odd, but (2n + r) is even.
In both cases, we have even * odd = 2^(x + 1).
The RHS is always 2*2*2* ..., none of which are odd.
Thus, there is a contradiction,
So powers of 2 cannot be trapezoidal numbers.

EDIT 2: On the number of addends -

2 addends: (m) + (m + 1)
= 2m + 1
= all odd numbers >= 3

3 addends: (m) + (m + 1) + (m + 2)
= 3m + 3 = 3(m + 1)
= all multiples of 3 >= 6

n addends: (m) + (m + 1) + (m + 2) + ... + [m + (n - 1)]
for a total of n terms.
= mn + (n - 1)n/2
= n(n + 2m - 1)/2 : This is the general formula.
Example: m = 7, n = 6
7 + 8 + 9 + 10 + 11 + 12 = 6(6 + 2*7 - 1)/2 = 57