Assuming that such an relationship exists, it must satisfy
S(1) = 1 = T(1) = a + b + c + d
S(2) = 5 = T(2) = 8a + 4b + 2c + d
S(3) = 14 = T(3) = 27a + 9b + 3c + d
S(4) = 30 = T(4) = 64a + 16b + 4c + d
4 = 7a + 3b + c
16 = 37a +7b + c
9 = 19a + 5b +c
12 = 30a + 4b
7 = 18a + 2b
14 = 36a + 4b
6a = 2
a = (1/3)
2b = 7 - 6 = 1
b = 1/2
c = 4 - 7a - 3b = 4 - 7/3 - 3/2 = 1/6
d = 1 -a -b -c = 0
T(n) = (1/3)n^3 + (1/2)n^2 + (1/6)n
Let's see if it satisfies T(5) = 55
(1/3)*125 + (1/2)*25 + (1/6)*5 = 55
It works for all n.
S(n) = 1+4+9+16+25+...+n^2
find the formula so it will look like T(n)=an^3+bn^2+cn+d and also find what a, b, c, d equals
2 answers
(sniff) I worked hard on that one.