question 1^2+4^2+7^2+10^2+...to n terms.

answer 1/2n(6n2-3n-1)

The base of your powers is an arithmetic sequence

sum(1) = 1
sum(2 ) = 1+ 4^2 = 17
sum(3) = 17 + 7^2 = 66
sum(4) = 66 + 10^2 = 166
sum(5) = 166 + 13^2 = 335
sum(6) = 335 + 16^2 = 591
...

with the third difference column of 18, we know it must be in the form of a cubic
we could set up 4 equations in 4 unknowns, as
y = ax^3 + bx^2 + cx + d
and plug in our first 4 points (1,1), (2,17), (3,66) and (4,335), such as
for (1,1)-- a+b+c+d = 1
for (2,17) 8a + 4b + 2c + d = 17
etc

or use Wolfram to do the drudgery for us:
http://www.wolframalpha.com/input/?i=pattern+%7B1,+17,+66,+166,+335,+591,+...%7D

factor out the n and you have your result

your sum is

n
∑ (3k-2)^2 = ∑9k^2 - 12∑k + ∑4
k=1

∑k^2 = n(n+1)(2n+1)/6
∑k = n(n+1)/2
∑4 = 4n

Add them up and you have

9n(n+1)(2n+1)/6 - 12n(n+1)/2 + 4n
= n(6n^2-3n-1)/2

To find the sum of the series 1^2 + 4^2 + 7^2 + 10^2 + ... up to n terms, we can use the formula for the sum of an arithmetic series.

In this series, each term can be written as (3n - 2)^2, where n is the position of the term.

The formula to find the sum of an arithmetic series is:
Sn = (n/2)(a + l)
where Sn is the sum of the series, n is the number of terms, a is the first term, and l is the last term.

Let's find the last term of the series. The nth term is given by:
(3n - 2)^2

So for the last term, we substitute n with n in the above expression:
l = (3n - 2)^2

Now, the sum of the series can be found using the formula:
Sn = (n/2)(a + l)

Substituting the values, we get:
Sn = (n/2)(1^2 + (3n - 2)^2)

Simplifying the expression,
Sn = (n/2)(1 + 9n^2 - 12n + 4)

Expanding further,
Sn = (n/2)(9n^2 - 12n + 5)

Simplifying the expression,
Sn = (9n^3 - 12n^2 + 5n) / 2

Therefore, the sum of the series 1^2 + 4^2 + 7^2 + 10^2 + ... up to n terms is given by the expression:
Sn = (9n^3 - 12n^2 + 5n) / 2

This can be further simplified to the answer you mentioned: 1/2n(6n^2 - 3n - 1).