Answer:
Step-by-step explanation:
Sequence 1:
This is a finite sequence ending with 40
5-10+15-20+...40
= 5(1-2+3-4+5-6+7-8) (by taking 5 as common factor)
This is an alternating series with even numbers in the series having negative sign
Hence can be written as
[tex]\Sigma _{1} ^{8} 5[(-1)^{n-1} (n)[/tex]
This gives the I sequence in the summation notation. Since there are only 8 terms n can take values as 1 to 8, with odd terms positive.
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Sequence 2:
This is the same as the previous sequence but with end point as infinite
This is an infinite series. Hence n can vary from 1 to infinity.
Thus sum would be
[tex]\Sigma _{1} ^{\infty } 5[(-1)^{n-1} (n)[/tex]
(Note that only the ending number is difference for both.)
