0
The sum of any two successive triangular number is a square number.
The nth triangular number is n(n+1)/2 = (n2 +n)/2
The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2
Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2
This is easy to visualise. For example T3 + t4 = 42
[T3, the 3th triangular number represented by X,
t4, the 4th triangular number represented by x]
XXX_____x
XX_____xx
X_____xxx
_____xxxx
Bring them together and you have:
XXXx
XXxx
Xxxx
xxxx
The sum of any two successive triangular number is a square number.
The nth triangular number is n(n+1)/2 = (n2 +n)/2
The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2
Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2
This is easy to visualise. For example T3 + t4 = 42
[T3, the 3th triangular number represented by X,
t4, the 4th triangular number represented by x]
XXX_____x
XX_____xx
X_____xxx
_____xxxx
Bring them together and you have:
XXXx
XXxx
Xxxx
xxxx
The sum of any two successive triangular number is a square number.
The nth triangular number is n(n+1)/2 = (n2 +n)/2
The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2
Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2
This is easy to visualise. For example T3 + t4 = 42
[T3, the 3th triangular number represented by X,
t4, the 4th triangular number represented by x]
XXX_____x
XX_____xx
X_____xxx
_____xxxx
Bring them together and you have:
XXXx
XXxx
Xxxx
xxxx
The sum of any two successive triangular number is a square number.
The nth triangular number is n(n+1)/2 = (n2 +n)/2
The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2
Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2
This is easy to visualise. For example T3 + t4 = 42
[T3, the 3th triangular number represented by X,
t4, the 4th triangular number represented by x]
XXX_____x
XX_____xx
X_____xxx
_____xxxx
Bring them together and you have:
XXXx
XXxx
Xxxx
xxxx
Still curious? Ask our experts.
Chat with our AI personalities
Beau
Lao
DevinThe sum of any two successive triangular number is a square number.
The nth triangular number is n(n+1)/2 = (n2 +n)/2
The next triangular number is (n+1)*(n+2)/2 = (n2 + 3n + 2)/2
Their sum is (2n2 + 4n + 2)/2 = n2 + 2n + 1 = (n + 1)2
This is easy to visualise. For example T3 + t4 = 42
[T3, the 3th triangular number represented by X,
t4, the 4th triangular number represented by x]
XXX_____x
XX_____xx
X_____xxx
_____xxxx
Bring them together and you have:
XXXx
XXxx
Xxxx
xxxx
Add your answer:
Earn +20 pts
What are the real numbers that is not a whole number?
Real numbers are all numbers which do not contain "i", when "i" represents the square root of -1. All numbers which do contain "i" are "imaginary numbers" and are not real numbers. This means that all numbers you'd ordinarily use are real numbers - all the counting numbers (integers) and all decimals are real numbers. So in answer to your question, all the real numbers that are not whole numbers are all the decimal numbers - including irrational decimals such as pi.
How do you check if a number is a triangular number?
you need to use this formula: n(n+1) T=--------- 2 So number times (number + 1) divided by 2. If the number you get is the same number as n its a triangular number. if it isn't well it isn't a triangular number.
There are numbers from 1 to n. they have to be inverted as n to 1 . How many moves are required if only two consecutive numbers are stampled at a time?
There are (n2 - n) / 2 moves required.I am not aware that "stampled" is a word beyond a cross between stampede and trample, but I believe you wish to reverse the order of numbers from 1 to n where only two consecutive numbers can be swapped per step.This process will show a pattern of triangular numbers. For example:Let us suppose there are 3 numbers 1,2,3: To move the last (highest) number to the first place requires 2 steps. The second highest number is now in the last place. To move it to the second place requires 1 step. Finished. Total = 3 steps.Or:If there are 4 numbers, i.e. 1,2,3,4:3 steps + 2 steps + 1 step = 6 steps total. We know it is only 3 steps more than the last example because once we have moved the 4 to the first place we then simply have to do the above again (rearrange the 1,2,3, to 3,2,1).So you can see that the number of steps are all triangular numbers.A triangular number is calculated by n(n+1) / 2. So we can use this formula, but we need to alter it because the number of steps are for the previous triangular number. E.g. Where there are 4 numbers we have to do 1+2+3 steps to reorder it (see above).Therefore if we make our formula (n - 1) n / 2 = (n2 - n) / 2 then this will work.
Where can you use prime numbers?
In every math problem you can ever do you can use prime numbers. 1+3 Both prime numbers. Most numbers are not prime but prime numbers only become a broblem when you try to factor them like in simple algebra.
What are similarities and differences between Mayan number system and Hindu Arabic number system?
Both make use of a zero symbol but Mayan numbers have 20 as a base whereas Hindu-Arabic numbers have 10 as a base.
