(1) This series consists of terms of an arithmetic sequence:
179 - 173 = 6
173 - 167 = 6
and so on, so that the n-th term in the series is (for n ≥ 1)
a(n) = 179 - 6 (n - 1) = 185 - 6n
Then the sum of the first 53 terms is
[tex]\displaystyle\sum_{n=1}^{53}(185-6n) = 185\sum_{n=1}^{53}1-6\sum_{n=1}^{53}n[/tex]
[tex]\displaystyle\sum_{n=1}^{53}(185-6n) = 185\times53-6\times\frac{53\times54}2[/tex]
[tex]\displaystyle\sum_{n=1}^{53}(185-6n) = \boxed{1219}[/tex]
(2) This series has terms from a geometric sequence:
-12 / 6 = -2
24/(-12) = -2
-48/24 = -2
and so on. The n-th term is (again, for n ≥ 1)
a(n) = 6 (-2)ⁿ⁻¹
and the sum of the first 19 terms is
[tex]\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 + (-2) + (-2)^2 + (-2)^3 + \cdots+(-2)^{19}\right)[/tex]
Multiply both sides by -2 :
[tex]\displaystyle-2\sum_{n=1}^{19}6(-2)^{n-1} = 6\left((-2) + (-2)^2 + (-2)^3 + (-2)^4 + \cdots+(-2)^{20}\right)[/tex]
Subtracting this from the first sum gives
[tex]\displaystyle(1-(-2))\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 -(-2)^{20}\right)[/tex]
and solving for the sum, you get
[tex]\displaystyle3\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 -(-2)^{20}\right)[/tex]
[tex]\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -(-2)^{20}\right)[/tex]
[tex]\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -(-1)^{20}2^{20}\right)[/tex]
[tex]\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -2^{20}\right) = 2-2^{21} = \boxed{-2,097,150}[/tex]
