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
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.
(sniff) I worked hard on that one.
To find the formula for S(n) in the desired form T(n) = an^3 + bn^2 + cn + d, we need to determine the values of a, b, c, and d.
To get started, let's consider the differences between consecutive terms in the series S(n):
1, 4, 9, 16, 25, ...
The differences are:
3, 5, 7, 9, ...
Notice that the differences form an arithmetic sequence. The first difference is 3, and each subsequent difference increases by 2. This suggests that the formula for the differences can be written as:
D(n) = 3 + 2(n-1) = 2n + 1
Where D(n) represents the nth difference.
Now, let's find an expression for S(n) in terms of the differences. To do this, we need to sum the differences up to the nth term:
S(n) = D(1) + D(2) + D(3) + ... + D(n)
By substituting the formula for D(n), we get:
S(n) = (2⋅1 + 1) + (2⋅2 + 1) + (2⋅3 + 1) + ... + (2⋅n + 1)
Simplifying further:
S(n) = 2⋅(1+2+3+...+n) + (1+1+1+...+1)
The sum of the first n natural numbers (1+2+3+...+n) can be expressed as n(n+1)/2. Substituting this, we have:
S(n) = 2⋅(n(n+1)/2) + (1⋅n)
Simplifying again:
S(n) = n(n+1) + n
S(n) = n^2 + n + n
Finally, we have the expression for S(n) in the form T(n) = an^3 + bn^2 + cn + d:
T(n) = n^2 + n^1 + n^0
Therefore, a = 1, b = 1, c = 1, and d = 0.