Problem 61
Refer to this link where I solved the problem earlier
https://brainly.com/question/24517029
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Problem 62
T(n) = nth triangular number
T(n) = n(n+1)/2 = 0.5n(n+1)
That squares to (0.5n(n+1))^2 = 0.25n^2(n+1)^2
The next triangular number after T(n) is
T(n+1) = 0.5(n+1)(n+1+1) = 0.5(n+1)(n+2)
That squares to 0.25(n+1)^2(n+2)^2
Notice how each squared result has 0.25 and (n+1)^2 found buried in them.
Let's say A = 0.25(n+1)^2
That would mean the first result 0.25n^2(n+1)^2 becomes An^2
The second result 0.25(n+1)^2(n+2)^2 becomes A(n+2)^2
Let's add those to see what happens
An^2+A(n+2)^2
An^2+A(n^2+4n+4)
A(n^2+n^2+4n+4)
A(2n^2+4n+4)
0.25(n+1)^2*(2(n^2+2n+2))
0.5(n+1)^2(n^2+2n+1+1)
0.5(n+1)^2((n+1)^2+1)
0.5k(k+1)
We see that the result is a triangular number where k = (n+1)^2
This shows that adding the squares of consecutive triangular numbers gets us another triangular number.
A few examples
- 3^2+6^2 = 9+36 = 45 which is in the form n(n+1)/2 when n = 9
- 6^2+10^2 = 36+100 = 136 which is in the form n(n+1)/2 when n = 16
Answer: Triangular number
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Problem 63
I'll borrow some of the ideas from problem 62
We found that after squaring the nth and (n+1)th triangular numbers, we got An^2 and A(n+2)^2 respectively. We let A = 0.5(n+1)^2
Subtract those expressions to get...
A(n+2)^2 - An^2
A(n^2+4n+4)-An^2
A(n^2+4n+4-n^2)
A(4n+4)
4A(n+1)
4*0.5(n+1)^2*(n+1)
(n+1)^3
This proves that the difference between the squares of consecutive triangular numbers is a perfect cube, aka a cube number.
A few examples:
- 6^2 - 3^2 = 36 - 9 = 27 = 3^3
- 10^2 - 6^2 = 100-36 = 64 = 4^3
Answer: Cube number
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Problem 64
Refer to this link where I solved the problem earlier
https://brainly.com/question/24517029
