Question

What is the 12th partial sum of the summation of negative 2i minus 10, from i equals 1 to infinity?

Answer 1

Ans: 12th partial sum = -276

Given :

`\sum_(i=1)^(12) (-2i-10)`

Find the values for i = 1 to 12:
x (1) = -2 – 10 = -12
x (2) = -4 – 10 = -14
x (3) = -6 – 10 = -16
x (4) = -8 – 10 = -18
x (5) = -10 – 10 = -20
x (6) = -12 – 10 = -22
x (7) = -14 – 10 = -24
x (8) = -16 – 10 = -26
x (9) = -18 – 10 = -28
x (10) = -20 – 10 = -30
x (11) = -22 – 10 = -32
x (12) = -24 - 10 = -34

Add all of them:
12th partial sum = -276