Proof by induction with variable other than $n$

Question

1) Prove that $(1+x)^{n} \geq 1 + nx$ for every $n \in \mathbb{N}$ and $x \in (-1, \infty)$

Base case: Usually for the base case I just take $n = 1$ but since there's another variable $x$, I wasn't sure which value to use for $x$, so I just set $x = -1$. Not sure if this is correct...

$(1 - 1)^{1} \geq 1 - 1$ which stands, so I proceed with the proof.

$n \rightarrow n + 1$: Assume $(1+x)^{n} \geq 1 + nx$ for every $n \in \mathbb{N}$ and $x \in (-1, \infty)$. I want to show that $(1 + x)^{n + 1} \geq 1 + (n + 1) x$ for every $n \in \mathbb{N}$ and $x \in (-1, \infty)$

$(1 + x)^{n} \cdot (1 + x) \geq (1 + nx) (1 + x)$ by the induction hypothesis

$\Rightarrow (1 + x)^{n + 1} \geq 1 + x + nx + nx^{2} \geq 1 + nx + x$

$ \square$

2) Prove that $\sum_{i = 1}^{n} i(i + 1) = n(n + 1)\frac{(n + 2)}{3}$ for all $n \in \mathbb{N}$

Base case: I get thrown off by $i$ (just like how I got throw off by $x$ in the previous problem), but I just set $i = 1$ and the base case stands:

$1( 1 + 1) = 1 \cdot 2 \cdot 1 \Rightarrow 2 = 2$

So I proceed with the proof.

$n \rightarrow n + 1$: Assume that $\sum_{i = 1}^{n} i(i + 1) = n(n + 1)\frac{(n + 2)}{3}$ for all $n \in \mathbb{N}$, want to show that...

$i(i + 1) = (n + 1)(n + 2) \frac{(n + 3)}{3}$ for all $n \in \mathbb{N}$

Is this correct? If so, wouldn't that imply $(n + 1)(n + 2) \frac{(n + 3)}{3} = n(n + 1)\frac{(n + 2)}{3}$? And that doesn't really make sense to me.

Answer 1

Your mistakes:

In the first step of 1) you have any $x>-1$, not $-1$, but for $n=1$ this is simply an identity $(1+x)^1=1+1\cdot{x}$.

In 2) you shold prove that $$ 1\cdot2+2\cdot3+\dots+n(n+1)=n(n + 1)\dfrac{(n + 2)}{3} $$

Answer 2

you must prove that $n(n+1)\left(\frac{n+2}{3}\right)+(n+1)(n+2)=(n+1)(n+2)\left(\frac{n+3}{3}\right)$