Question

1. Determine the sum of the first 53 terms of the following series: 179+173+167+...

2. Determine the sum of the first 19 terms of the following series: 6−12+24−48+...

Answer 1

(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

`\displaystyle\sum_(n=1)^(53)(185-6n) = 185\sum_(n=1)^(53)1-6\sum_(n=1)^(53)n`

`\displaystyle\sum_(n=1)^(53)(185-6n) = 185*53-6*\frac{53*54}2`

`\displaystyle\sum_(n=1)^(53)(185-6n) = \boxed{1219}`

(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

`\displaystyle\sum_(n=1)^(19)6(-2)^(n-1) = 6\left(1 + (-2) + (-2)^2 + (-2)^3 + \cdots+(-2)^(19)\right)`

Multiply both sides by -2 :

`\displaystyle-2\sum_(n=1)^(19)6(-2)^(n-1) = 6\left((-2) + (-2)^2 + (-2)^3 + (-2)^4 + \cdots+(-2)^(20)\right)`

Subtracting this from the first sum gives

`\displaystyle(1-(-2))\sum_(n=1)^(19)6(-2)^(n-1) = 6\left(1 -(-2)^(20)\right)`

and solving for the sum, you get

`\displaystyle3\sum_(n=1)^(19)6(-2)^(n-1) = 6\left(1 -(-2)^(20)\right)`

`\displaystyle\sum_(n=1)^(19)6(-2)^(n-1) = 2\left(1 -(-2)^(20)\right)`

`\displaystyle\sum_(n=1)^(19)6(-2)^(n-1) = 2\left(1 -(-1)^(20)2^(20)\right)`

`\displaystyle\sum_(n=1)^(19)6(-2)^(n-1) = 2\left(1 -2^(20)\right) = 2-2^(21) = \boxed{-2,097,150}`