Find the 12th partial sum of the summation of negative 2i minus 10 from i equals 1 to infinity

To find the 12th partial sum of the series given by the summation of (-2i – 10) from i equals 1 to 12, we first need to establish the formula for the partial sum.

The formula for the partial sum S_n of a series where the terms are defined by a function f(i) is:

S_n = Σ (from i=1 to n) f(i)

Here, f(i) = -2i – 10.

Now, we will calculate S₁₂:

S₁₂ = Σ (from i=1 to 12) (-2i – 10)

Breaking this down:

S₁₂ = Σ (from i=1 to 12)(-2i) + Σ (from i=1 to 12)(-10)

The first part, Σ (from i=1 to 12)(-2i), can be simplified:

Σ (from i=1 to 12)(-2i) = -2Σ (from i=1 to 12)(i)

We know that Σ (from i=1 to n)(i) = n(n + 1)/2. Therefore, Σ (from i=1 to 12)(i) = 12(12 + 1)/2 = 12 * 13 / 2 = 78.

Thus, -2Σ (from i=1 to 12)(i) = -2 * 78 = -156.

Next, we consider the second part, Σ (from i=1 to 12)(-10):

Σ (from i=1 to 12)(-10) = -10 * 12 = -120.

Putting it all together:

S₁₂ = -156 – 120 = -276.

Therefore, the 12th partial sum of the series is:

-276