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Ipatiy [6.2K]
3 years ago
10

IF I GET AN ANSWER IN 10 MIN UR VOTED BRAINLIEST

Mathematics
1 answer:
Over [174]3 years ago
3 0

Adding and subtracting with negative numbers

First, recognize that adding and subtracting are, from one viewpoint, the same thing. Subtracting a number is the same thing as adding the negative of that number. For example, 4 – 12 is the same as 4 + –12 (which, because the order of terms doesn't matter with addition, is the same as –12 + 4). With that in mind, here are the rules for adding with negative and positive numbers:

If both numbers are positive, then the answer is positive.

If both numbers are negative, then the answer is negative.

If the numbers have different signs, the answer takes the sign of the higher number.

Subtracting a negative number is the same as adding the positive of that number. For example, 5 – –4 is the same as 5 + 4.

Multiplying and dividing with negative numbers if the numbers have different signs, the answer is negative.

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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
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\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
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\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
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