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3 years ago

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 >/ 1

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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.

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3 years ago