Justify the last two steps of the proof. Help pls :)
1 answer:
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<h3>Answer: C) 6</h3>
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Explanation:
The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.
It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.
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If n = 1, then we have
(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5
Let A = -5 as we'll use it later.
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If n = 2, then
(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8
Let B = 8 since we'll use this later as well
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If n = 3, then
(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11
Let C = -11
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If n = 4, then
(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14
Let D = 14.
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We'll add up the values of A,B,C,D to get the final answer
A+B+C+D = -5+8+(-11)+14 = 6
This means that