Please try this, it's a basic proof that I don't know how to do. Thanks!
1 answer:
A cube, is made off 6 squarial faces, so all faces on that cube, are squares, the front, back, left, right, top and bottom.
a square has all equal sides, and also all right angles, so all angles in a square are 90°. Let's say the sides are "x" long.
now, if we run a plane on that cube diagonally, check the picture below, the diagonal side at the bottom, by usin the 45-45-90 rule as you see it there, will be x√2.
let's keep in mind that, "x" is opposite side of that angle θ, and then x√2 will be the adjacent side of it.
and we can use those two to get the tangent and then the inverse tangent to get the value, as you see it in the picture.
if you need the angle in radians, run the inverse tangent again, just make sure your calculator is in radians mode.
a square has all equal sides, and also all right angles, so all angles in a square are 90°. Let's say the sides are "x" long.
now, if we run a plane on that cube diagonally, check the picture below, the diagonal side at the bottom, by usin the 45-45-90 rule as you see it there, will be x√2.
let's keep in mind that, "x" is opposite side of that angle θ, and then x√2 will be the adjacent side of it.
and we can use those two to get the tangent and then the inverse tangent to get the value, as you see it in the picture.
if you need the angle in radians, run the inverse tangent again, just make sure your calculator is in radians mode.
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Problem 61
The nth triangular number is
T(n) = n(n+1)/2
I'll rewrite this into
T(n) = 0.5n(n+1)
The triangular number right after this is
T(n+1) = 0.5(n+1)(n+2)
I replaced every n with n+1 and simplified
Let's see what we get when we add up the two expressions
T(n) + T(n+1)
0.5n(n+1) + 0.5(n+1)(n+2)
0.5n^2+0.5n + 0.5(n^2+3n+2)
0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1
n^2+2n+1
(n+1)^2
This shows that the sum of any two consecutive triangular numbers results in a square number
Here's a few examples
- 0+1 = 1
- 1+3 = 4
- 3+6 = 9
- 6+10 = 16
- 10+15 = 25
Note each sum is a perfect square, which visually would plot out a square figure.
For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}
<h3>Answer: Square number</h3>==========================================================
Problem 64
Let's say we go with n = 5.
This means,
T(n) = 0.5n(n+1)
T(n-1) = 0.5(n-1)(n-1+1)
T(n-1) = 0.5n(n-1)
T(5-1) = 0.5*5(5-1)
T(4) = 10
This says that when n = 5, the 4th triangular number is 10
Triple that result and add on n = 5
3*T(4) + n = 3*10+5 = 35
This result is beyond obvious which category of figurate number it belongs to. It's not a triangular number since it's not in the form n(n+1)/2. It's not a square number either.
Through a bit of trial and error, you should find it's a pentagonal number
Pentagonal numbers are of the form n(3n-1)/2
If you plugged n = 5 into that, it leads to 35
n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35
The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)
<h3>Answer: Pentagonal number</h3>