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
Explanation:
The
#T_n = 1/2 n(n+1)#
They start:
#1, 3, 6, 10, 15,...#
If
#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
#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
#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#