(Im not English and just started doing maths in English so my termiology is still way off)
So the title for $n\ge 3$
First I use calculate both sides with $3$, which is true
I make my induction. $(k+1)^2 < 2k^2$
then I replace $N$ with $k+1$: $(k+2)^2 < 2(k+1)^2$
Now what? I cant seem to find how to use my induction in this one. I've also tried working out the brackets, but that also didn't seem to help me.
HINT: $(k+2)^2=\big((k+1)+1\big)^2=(k+1)^2+2(k+1)+1$; now apply the induction hypothesis that $(k+1)^2<2k^2$. (There will still be a bit of work to do; in particular, you’ll have to use the fact that $k\ge 1$.)
((2.)) (k+1)^2 + 2(k+1) + 1 < 2k^2 + 4k + 2
((3.)) (k+1)^2 + 2k + 2 < 2k^2 + 4k + 2
((4.)) then my hypothesis says (k+1)^2 < 2k^2 and 2k+2 < 4k+2
right?
– Wouter Zirkzee Sep 07 '13 at 19:18$$(n+1)^2<2n^2\iff \left(1+\frac1n\right)^2<2$$
Let $P(n): \left(1+\frac1n\right)^2<2$
For $n=3,\left(1+\frac1n\right)^2=\left(1+\frac13\right)^2=\frac{16}9<2$
Let $P(n)$ is true for $n=m\ge3$ i.e., $\left(1+\frac1m\right)^2<2$
Now, as $m+1>m\iff \frac1m>\frac1{m+1}$ $\displaystyle \implies \left(1+\frac1{m+1}\right)^2<\left(1+\frac1m\right)^2<2$
So, $P(m+1)$ is true if $P(m)$ is true
1.Basis
First we check for the smallest number: $n=3$
$$4^2 < 2*3^2$$ $$16 < 18$$
2.Inductive Hypothesis
$$(k+1)^2 < 2k^2$$
3.Inductive step
$$(k+2)^2 < 2(k+1)^2$$ $$k^2 + 4k + 4 < 2k^2 + 4k + 2$$ $$4 < k^2 + 2$$
Which is true for every $n \geq 3$
$(k+2)^2=(k+1)^2+2k+3$ and $2(k+1)^2=2k^2+4k+2$. Then by inductive hypothesis, $(k+1)^2<2k^2$, and for $k>0$, $2k+3<4k+2$. Thus you get $(k+2)^2<2(k+1)^2$.
