please can u tell me the term to term rule to this question 0,1,3,6,10,15,21 {nth}
thank u
Those are the triangular numbers, except your sequence started with 0 and we usually start with 1.
0= 1*(11)/2 term 1
1= 2*(21)/2 term 2
3= 3*(31)/2 term 3
6= 4*(41)/2 term 4
10=5*(51)/2 term 5
15=6*(61)/2 term 6
.
.
.
Do you see a progression? What is the nth term?
nth term is like if the 0,1,3,6,10,15,21 is not there we have to use n as a number
so do know how i can answer this question please
thanks
Sort of...what we have is a sequence. In a sequence we match each term with the positive integers like this
1 0
2 1
3 3
4 6
5 10
6 15
.
.
.
n ??
The nth term will some kind of expression so that when we put the number n into it we'll get the value of that term in the sequence.
Here's an example
1 3
2 6
3 9
.
.
.
n 3n
So the nth term of the sequence 3,6,9... would be 3*n
Now study the sequnce I wrote in the previous post here and see where n should be used.
Triangular Numbers
The number of dots, circles, spheres, etc., that can be arranged in an equilateral or right triangular pattern is called a triangular number. The 10 bowling pins form a triangular number as do the 15 balls racked up on a pool table. Upon further inspection, it becomes immediately clear that the triangular numbers, T1, T2, T3, T4, etc., are simply the sum of the consecutive integers 1234.....n or Tn = n(n + 1)/2, namely, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66,78, 91, etc.
Triangular numbers are the sum of the balls in the triangle as defined by Tn = n(n + 1)/2.
Order.n...1........2...............3.....................4.............................5....................6.....7.....8.....9
.............O.......O...............O....................O............................O
....................O...O.........O...O...............O...O.......................O...O
..................................O...O...O.........O....O....O...............O....O....O
......................................................O....O...O....O.........O.....O...O....O
.................................................................................O....O....O....O....O
Total......1........3.................6....................10..........................15..................21...28...36...45...etc.
The sum of a series of triangular numbers from 1 through Tn is given by S = (n^3 + 3n^2 + 2n)/6.
After staring at several triangular and square polygonal number arrangements, one can quickly see that the 1st and 2nd triangular numbers actually form the 2nd square number 4. Similarly, the 2nd and 3rd triangulars numbers form the 3rd square number 9, and so on. By inspection, one can see that the nth square number, Sn, is equal to Tn + T(n  1) = n^2. This can best be visualized from the following:
.........Tn  1...3...6...10...15...21...28...36...45...55...66...78...91
.........T(n  1)........1...3....6....10...15...21...28...36...45...55...66...78
.........Sn.........1...4...9....16...25...36...49...64...81..100.121.144.169
A number cannot be triangular if its digital root is 2, 4, 5, 7 or 8.
Some interesting characteristice of Triangular numbers:
The numbers 1 and 36 are both square and triangular. Some other triangular squares are 1225, 41,616, 1,413,721, 48,024,900 and 1,631,432,881. Triangular squares can be derived from the series 0, 1, 6, 35, 204, 1189............Un where Un = 6U(n  1)  U(n  2) where each term is six times the previous term, diminished by the one before that. The squares of these numbers are simultaneously square and triangular.
The difference between the squares of two consecutive rank triangular numbers is equal to the cube of the larger numbers rank.
Thus, (Tn)^2  (T(n  1))^2 = n^3. For example, T6^2  T5^2 = 441  225 = 216 = 6^3.
The summation of varying sets of consecutive triangular numbers offers some strange results.
T1 + T2 + T3 = 1 + 3 + 6 = 10 = T4.
T5 + T6 + T7 + T8 = 15 + 21 + 28 + 36 = 100 = 45 + 55 = T9 + T10.
The pattern continues with the next 5 Tn's summing to the next 3 Tn's followed by the next6 Tn's summing to the next 4 Tn's, etc.
The sum of the first "n" cubes is equal to the nth triangular number. For instance:
n............1.....2.....3.....4.......5
Tn..........1.....3.....6....10.....15
n^3.........1 + 8 + 27 + 64 + 100 = 225 = 15^2
Every number can be expressed by the sum of three or less triangular numbers, not necessarily different.
1 = 1, 2 = 1 + 1, 3 = 3, 4 = 3 + 1, 5 = 3 + 1 + 1, 6 = 6, 7 = 6 + 1, 8 = 6 + 1 + 1, 9 = 6 + 3, 10 = 10, etc.
Alternate ways of finding triangular squares.
From Tn = n(n + 1)/2 and Sn = m^2, we get m^2 = n(n + 1)/2 or 4n^2 + 4n = 8m^2.
Adding one to both sides, we obtain 4n^2 + 4n + 1 = 8m^2 + 1.
Factoring, we find (2n + 1)^2 = 8m^2 + 1.
If we allow (2n + 1) to equal "x" and "y" to equal 2m, we come upon x^2  2y^2 = 1, the famous Pell Equation.
We now know that the positive integer solutions to the Pell equation, x^2  2y^2 = +1 lead to triangular squares. But how?
Without getting into the theoretical aspect of the subject, sufficeth to say that the Pell equaion is closely connected with early methods of approximating the square root of a number. The solutions to Pell's equation, i.e., (x,y), often written as (x/y) are approximations of the square root of D in x^2  2y^2 = +1. Numerous methods have evolved over the centuries for estimating the square root of a number.
Diophantus' method leads to the minimum solutions to x^2  Dy^2 = +1, D a non square, by setting x = my + 1 which leads to y = 2m/(D  m^2).
From values of m = 1.......n, many rational solutions evolve.
Eventually, an integer solution will be reached.
For instance, the smallest solution to x^2  2y^2 = +1 derives from m = 1 resulting in x = 3 and y = 2 or sqrt(2) ~= 3/2..
Newton's method leads to the minimum solution sqrt(D) = sqrt(a^2 + r) = (a + D/a)/2 ("a" = the nearest square) = (3/2).
Heron/Archimedes/El Hassar/Aryabhatta obtained the minimum solution sqrt(D) = sqrt(a^2 +r) = a +r/2a = (x/y) = (3/2).
Other methods exist that produce values of x/y but end up being solutions to x^2  Dy^2 = +/C.
Having the minimal solutions of x1 and y1 for x^2  Dy^2 = +1, others are derivable from the following:
(x + ysqrtD) = (x1 + y1sqrtD)^n, n = 1, 2, 3, etc.
Alternitive approach
Given x = p and y = q satifying x^2  2y^2 = +1, we can write (x + sqrtD)(x  sqrtD) = 1.
x = [(p + qsqrt(2))^n + (p  qsqrt(2))^n]/2
y = [(p + qsqrt(2))^n  (p  qsqrt(2))^n]/(2sqrt(2))
Having the minimum solution of x = 3 and y = 2, the next few solutions derive from n = 2 and 3 where x = 17, y = 12, x = 99 and y = 70 respectively.
Alternative approach
Subsequent solutions can also be obtained by means of the following:
x^2  2y^2 = +1 can be rewritten as x^2  2y^2 = (x + yqrt(2)(x  ysqrt(2)) = +1.
Using the minimum solution of x = 3 and y = 2, we can now write
.................(3 + sqrt(2))^2(3  sqrt(2))^2 = 1^2 = 1
.................(17 + 12sqrt(2))(17  12sqrt(2)) = 1
.................289  2(144) = 17^2  2(12)^2 = 1 the next smallest solution.
The next smallest solution is derivable from
.................(3 + sqrt(2))^3(3  sqrt(2))^3 = 1^2 = 1 which works out to
.................(99 + 70sqrt(2))(99  70sqrt(2)) = 1 or
.................99^2  2(70)^2 = 1.
Similarly, (3 + sqrt(2))^4(3  sqrt(2))^4 = 1^2 = 1 leads to
.................(577 + 408sqrt(2))(577  408sqrt(2)) = 1 and
.................577^2  2(408)^2 = 1.
Regardless of the method, we ultimately end up with the starting list of triangular squares.
..x........y........n.......m........Tn = Sm^2
..3........2........1........1..............1
.17......12.......8........6..............36
.99......70......49......35...........1225
577....408....288.....204.........41,616 etc.
3=4(65)+23
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