The given solution is not correct...
Let a(n) be the n-th term in the sequence. Since a(n) is quadratic, we know that we can write
a(n) = b n² + c n + d
for some constants b, c, d.
Using the given known values of a(n), we then have
a(1) = b + c + d = -3
a(2) = 4b + 2c + d = 6
a(3) = 9b + 3c + d = 27
Solve these equations for b, c, and d.
• Eliminate d from the first two equations:
(4b + 2c + d) - (b + c + d) = 6 - (-3) ⇒ 3b + c = 9
(9b + 3c + d) - (b + c + d) = 27 - (-3) ⇒ 8b + 2c = 30
• Eliminate c from the two new equations and solve for b :
(8b + 2c) - 2 (3b + c) = 30 - 2(9) ⇒ 2b = 12 ⇒ b = 6
• Solve for c and d :
3b + c = 3(6) + c = 9 ⇒ c = -9
b + c + d = 6 - 9 + d = -3 ⇒ d = 0
So, a(n) = 6n² - 9n.
The sum of the first n terms of this sequence is
[tex]\displaystyle \sum_{i=1}^n a(i) = \sum_{i=1}^n (6i^2 - 9i)[/tex]
Recall the formulas for sums of powers,
[tex]\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6[/tex]
Then the sum we want is
[tex]\displaystyle \sum_{i=1}^n (6i^2 - 9i) = 6\cdot\frac{n(n+1)(2n+1)}6 - 9\cdot\frac{n(n+1)}2 = \boxed{\frac{n(n+1)(4n-7)}2}[/tex]
