Assistance needed plz (1 Viewer)

HeroicPandas · Jan 16, 2013

I forgot how to do induction involving inequlities...

$$(i) Solve for x^2 > 2x + 1 \\ \\ $ $ (ii)Prove by mathematical induction that 2^n > n^2 $ $ for all integers n \geq 5$

My working out
$(i) x > 1 +\sqrt2 $ $ or$ $ x < 1-\sqrt2$

$$ Step 1: n =5 \\ \\ $ $\therefore 32 > 25, $ $\therefore$ $ true for n =5$

$$ Step 2: Assume true for n = k \\ \\$ $ ie. 2^k > k^2$

$Step 3: RTP: n = k + 1 \\ \\$ $ 2^{k+1}>(k+1)^2 \\ \\$ $2.2^k - (k^2 + 2k + 1) >0 \\ \\$$

What do i do now..?

Trebla · Jan 16, 2013

HeroicPandas said:
I forgot how to do induction involving inequlities...

$$(i) Solve for x^2 > 2x + 1 \\ \\ $ $ (ii)Prove by mathematical induction that 2^n > n^2 $ $ for all integers n \geq 5$

My working out
$(i) x > 1 +\sqrt2 $ $ or$ $ x < 1-\sqrt2$

$$ Step 1: n =5 \\ \\ $ $\therefore 32 > 25, $ $\therefore$ $ true for n =5$

$$ Step 2: Assume true for n = k \\ \\$ $ ie. 2^k > k^2$

$Step 3: RTP: n = k + 1 \\ \\$ $ 2^{k+1}>(k+1)^2 \\ \\$ $2.2^k - (k^2 + 2k + 1) >0 \\ \\$$

What do i do now..?

$Required to prove $\\\\2^{k+1} > (k + 1)^2\\\\$LHS $ = 2^{k+1}\\\\=2.2^{k}\\\\>2k^2\quad$ by assumption$\\\\=k^2 + k^2\\\\> k^2 + 2k + 1\quad$ from the previous part$\\\\=(k+1)^2\\\\=$ RHS$

HeroicPandas · Jan 16, 2013

Trebla said:
$Required to prove $\\\\2^{k+1} > (k + 1)^2\\\\$LHS $ = 2^{k+1}\\\\=2.2^{k}\\\\>2k^2\quad$ by assumption$\\\\=k^2 + k^2\\\\> k^2 + 2k + 1\quad$ from the previous part$\\\\=(k+1)^2\\\\=$ RHS$

Thank you so much!

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Assistance needed plz (1 Viewer)

HeroicPandas

Heroic!

Trebla

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HeroicPandas

Heroic!

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