One relationship is that the sum of the nth and the previous triangular numbers is equal to the nth square number.That isT(n-1) + T(n) = S(n)where T(n) is the nth triangular number and S(n) is the nth square number.
If your calculator has an exponentiation function, simply raise the number to the power of .5 Remember this trick: the nth root of X = X ^ (1/n)
The nth triangulat number is n*(n+1)/2 The 100th is 100*101/2 = 5050
The nth even number is 2n...
The nth triangulat number is n*(n+1)/2. So, the 61st is 61*62/2 = 1891
One relationship is that the sum of the nth and the previous triangular numbers is equal to the nth square number.That isT(n-1) + T(n) = S(n)where T(n) is the nth triangular number and S(n) is the nth square number.
The nth triangular number is n(n+1)/2
It means to find the generic expression for any term, not just for one specific term. For example, in the sequence 1, 4, 9, 16, 25... it should be clear that these are square numbers, which you can write as 12, 22, 32, etc. The nth. term is n2, meaning that for any number "n", term number n is n2.
If your calculator has an exponentiation function, simply raise the number to the power of .5 Remember this trick: the nth root of X = X ^ (1/n)
The nth triangulat number is n*(n+1)/2 The 100th is 100*101/2 = 5050
The radical symbol, otherwise known as the "square root sign", lets you take the nth root of any number.Any number can be placed above, and slightly to the left, of the square root sign, to indicate the nth root. For example, the cube root of 27 is 3.The number inside the square root sign (that which you are finding the square root of), is called the radicand.
You seem to be unaware of the fact that you can obtain the answer easily by using the scientific calculator that comes as part of your computer. In general the nth root is extremely difficult to find.
The nth triangular number is n(n+1)/2
The nth even number is 2n...
The nth triangulat number is n*(n+1)/2. So, the 61st is 61*62/2 = 1891
The sum of the first n cubed numbers is the square of the nth triangular number.
Given any number you want for the nth term (n>4), it is possible to find a polynomial of order 4 that will fit these points and the additional one. Consequently, any number can be the nth number - the rule for that is easy to find. Still, the simplest solution here is a polynomial of degree 1: Un = 0.8n + 2