Video Transcript
Express the series 496 minus 497
plus 498 minus 499 and so on, all the way up to minus 531, in sigma notation.
We notice here that the terms in
our series alternate from positive to negative. A series can be written using sigma
notation as shown: the sum of the general term π’ sub π where π takes values from
one to π. In this question, we need to find
an expression for this general term π’ sub π as well as the value of π. Letβs begin by considering the
sequence of positive integers 496, 497, 498, and so on, all the way up to 531. We can rewrite each of these terms
as the constant 495 plus some integer. 496 is 495 plus one, 497 is 495
plus two, and so on, all the way up to 531 which is equal to 495 plus 36. This suggests that we have 36 terms
in our sequence.
If we were looking to write this
series in sigma notation, we would have the sum of 495 plus π where π takes values
from one to 36. The series in this question is
slightly more complicated, though, as we want the second, fourth, sixth, and every
other even term to be negative. We can make any positive integer
negative by multiplying by negative one. If we multiplied each of our terms
by negative one to the power of π, the first, third, fifth, and all our terms would
be negative. As we want the even terms to be
negative, we need to multiply by negative one to the power of π plus one. The series 496 minus 497 plus 498
and so on can be written in sigma notation as negative one to the power of π plus
one multiplied by π plus 495, where π takes values from one to 36.
