0

This is from the question proof by induction

This is for $ n \ge4 $

$$ 2^{n+1} = 2^n*2 > 2n^2 = n^2 + n^2 > n^2 + 2n + 1 = (n + 1)^2 $$

$$2^{n+1}>(n + 1)^2$$

The part that I can't understand is how they got this part $$ n^2 + n^2 > n^2 + 2n + 1$$

user98761
  • 407

4 Answers4

4

Proving the step $n^2+n^2\geq n^2+2n+1$

$$\begin{array}{rl|r}n^2+n^2&= n^2+n\cdot n&\text{by properties of exponents}\\ &> n^2 +3n&\text{since}~n>3\\ &=n^2+2n+n&\text{by properties of addition}\\ &>n^2+2n+1&\text{since}~n>1\end{array}$$

JMoravitz
  • 80,908
  • Thank you so much for your answer! I just have one question, how come we're allowed to use the fact $n>1$ if we put the limitation that the inequality holds only for all integer n greater than 3? – user98761 Nov 05 '17 at 02:55
  • 1
    $n>3$ and $3>1$ together imply that $n>1$ is true as well. Any conclusion that follows from hypotheses can further be used later in the proof to prove something else. This is despite that the inequality $n>1$ is not strict, it is still a useful inequality for us here and is enough to imply what we want for that particular step. – JMoravitz Nov 05 '17 at 02:57
0

It amounts to $n^2>2n+1\iff n(n-2)>1$, and as $n-2\ge 2$...

Bernard
  • 179,256
0

Subtract $n^2+2n+1$ from both sides to get $n^2-2n-1>0$. Solve the inequality for $n$. You should find that for $n>4$, The inequality is true.

AlgorithmsX
  • 4,624
0

We know this is true for a value of $n$. We must now prove it for $n+1$

\begin{align} 2^{n+1} &\gt (n+1)^2 &\qquad\text{This is what we want to prove}\\ 2^{n+1} &= 2^n \cdot2 &\qquad\text{This is true for all n}\\ 2^n &> n^2 &\qquad\text{This is true by induction}\\ 2^n \cdot 2&> 2n^2 &\qquad\text{Multiplication}\\ 2^n \cdot 2&> n^2 + n^2 &\qquad\text{Just splitting it up}\\ n^2 + n^2 &> n^2 + 2n + 1&\qquad\text{We're trying to get to the first formula }\\ n^2 &\gt 2n+1 &\quad\text{True for all values of $n$ that satisfy the original induction}\\ n^2 + n^2 &> (n+1)^2 &\qquad\text{And now we simplify}\\ \end{align} $$2^{n+1} \gt 2n^2 \gt (n+1)^2$$

John Lou
  • 2,436
  • We are trying to get the first formula, so we want to manipulate one side to give us $(n+1)^2$. Clearly, we can't get to it from an exponential equation, so we switch to comparing the two quadratics. The formula in itself is obviously true for $n \gt 3$, so it will work for all the $n$ we pick. The key is to realize that getting this inequality will help us get the first inequality that I have. – John Lou Nov 05 '17 at 02:45