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lakkis [162]
3 years ago
5

The integer -1 is part of the solution set of x > -2. True False

Mathematics
1 answer:
Mumz [18]3 years ago
3 0
The answer is True.
The answer is True.
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Solve the following by using mathematical Induction. For &gt;/ 1
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Answer:

See explanation

Step-by-step explanation:

Prove that

1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)

1. When n=1, we have

  • in left part 1^2=1;
  • in right part \dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.

2. Assume that for all k following equality is true

1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)

3. Prove that for k+1 the following equality is true too.

1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)

Consider left part:

1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}

Consider right part:

\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)

We get the same left and right parts, so the equality is true for k+1.

By mathematical induction, this equality is true for all n.

3 0
3 years ago
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