What is the formula for the #n#th term #t_n# of the sequence #1^2, 1^3, 2^3, 1^3, 2^3, 3^3, 1^3, 2^3, 3^3, 4^3,...# ?

1 Answer
Nov 27, 2017

#t_n = (n-1/2 floor(1/2(sqrt(8n-7)-1)) (floor(1/2(sqrt(8n-7)-1)) + 1))^3#

Explanation:

The #n#th triangular number #T_n# is given by the formula:

#T_n = 1/2 n(n+1)#

They start:

#1, 3, 6, 10, 15,...#

If #1/2 n(n+1) = t# then:

#0 = 8(1/2 n(n+1) - t)#

#color(white)(0) = 4n^2+4n-8t#

#color(white)(0) = 4n^2+4n+1-(8t+1)#

#color(white)(0) = (2n+1)^2-(sqrt(8t+1))^2#

#color(white)(0) = (2n+1-sqrt(8t+1))(2n+1+sqrt(8t+1))#

Hence the positive value of #n# is:

#1/2(sqrt(8t+1)-1)#

Write down the sequence of positive natural numbers:

#1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21...#

Write down the sequence of cube roots of the desired sequence:

#1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6,...#

Subtract this sequence from the first sequence to get:

#0, 1, 1, 3, 3, 3, 6, 6, 6, 6, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 15,.... #

These are triangular numbers with indices:

#0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5,...#

The formula for a term of this last sequence can be written:

#a_n = floor(1/2(sqrt(8(n-1)+1)-1))#

#color(white)(a_n) = floor(1/2(sqrt(8n-7)-1))#

So the formula for the previous sequence can be written:

#b_n = T_(a_n) = 1/2 floor(1/2(sqrt(8n-7)-1)) (floor(1/2(sqrt(8n-7)-1)) + 1)#

Then subtracting this from #n# we get the sequence:

#c_n = n-1/2 floor(1/2(sqrt(8n-7)-1)) (floor(1/2(sqrt(8n-7)-1)) + 1)#

#1, 1, 2, 1, 2, 3, 1, 2, 3, 4,...#

Then we just need to cube this to get the desired sequence:

#t_n = (n-1/2 floor(1/2(sqrt(8n-7)-1)) (floor(1/2(sqrt(8n-7)-1)) + 1))^3#