- #1

- 598

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## Homework Statement

[tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx

## Homework Equations

## The Attempt at a Solution

I want to show for P(k+1)

(1+x)^(k+1)[tex]\geq[/tex]1+kx+x

(1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x

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- Thread starter kathrynag
- Start date

- #1

- 598

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[tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx

I want to show for P(k+1)

(1+x)^(k+1)[tex]\geq[/tex]1+kx+x

(1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x

- #2

- 179

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(1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x

Use this:

[tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx

- #3

- 598

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I don't really understand how I use that...

- #4

- 179

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An inequality still holds if you multiply both sides by a positive number. If you multiply both sides of an inequality by a negative number, then you have to flip the sign from [tex]\le[/tex] to [tex]\ge[/tex] or vice versa.

- #5

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Ok, so:

(1+x)^k(1+x)>(1+kx)(1+x)

>(1+2kx+x)

(1+x)^k(1+x)>(1+kx)(1+x)

>(1+2kx+x)

- #6

- 179

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Does the question give any restrictions on x? Is what you wrote still true of 1 + x < 0?

- #7

- 598

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No restrictions.

1+kx+x+kx^2

1+kx+x+kx^2

- #8

- 179

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[tex](1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx[/tex]

which makes the statement false.

- #9

- 598

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So, we have to assume 1+x>0

- #10

- 598

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so, x>-1

- #11

Mark44

Mentor

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I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.So, we have to assume 1+x>0

- #12

- 598

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No restrictions.

1+kx+x+kx^2

Once I get here I'm unsure where to go

- #13

- 179

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Once I get here I'm unsure where to go

Look at your first post; you need to show that that is [tex]\ge 1 + kx + x[/tex].

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