X-intercept = -3, y-intercept = 4A, B, C, or D
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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>