Question 7.

Finding the terms in the sequence, we see that $t_3=0$, $t_4=-\dfrac{1}{3}$, $t_5=0$

We would notice that all the odd terms are 0, and we are also asked the sum of only even terms, so we do not need to consider those

$t_6=-\dfrac{1}{5}$

We see that the even terms are in an HP: $-1,\ -\dfrac{1}{3},\ -\dfrac{1}{5},\ -\dfrac{1}{7},\ ...$

The sum we are asked is the inverse of these terms, that is: -1, -3, -5, -7, up to 1012 terms

The sum of this AP would be $\dfrac{\left[-\left(2\times\ 1\right)+\left(1012-1\right)\left(-2\right)\right]}{2}\times\ 1012$

Which is equal to $-1012\times\ 1012\ =\ -1024144$

Therefore, Option A is the correct answer.