Proof for maths induction

Question

prove: $$ \sum_{r=1}^{n}[r^{2}+1](r!)=n[(n+1)!]$$ for all $n \in N$

prove $n=1$, $(1^2+1)(1!)$ = $1[(1+1)!]$

assume true for $n=k$, $(k^2+1)$$(k!)$= $k$$[(k+1)!]$

I got to here : $[k^{2}+1](k!)+[(k+1)^2+1](k+1)! = (k+1)[(k+2)!]$

Answer 1

The inductive step is:

$$\sum_{r=1}^{n+1}(r^2+1)r!=\sum_{r=1}^n(r^2+1)r!+\left[(n+1)^2+1\right](n+1)!\stackrel{\text{Ind. Step}}=$$

$$=n\color{red}{(n+1)!}+\left[(n+1)^2+1\right]\color{red}{(n+1)!}=\color{red}{(n+1)!}\left[n+(n+1)^2+1\right]=$$

$$=(n+1)!\left[(n+1)(n+2)\right]=(n+1)(n+2)!\;\ldots$$

Answer 2

If $$ \sum_{r=0}^{n}(r^{2}+1)r!=n(n+1)!$$ then $$ \sum_{r=0}^{n+1}(r^{2}+1)r!=\sum_{r=0}^{n}(r^{2}+1)r!+((n+1)^2+1)(n+1)!=$$ $$=n(n+1)!+((n+1)^2+1)(n+1)!=$$ $$=(n+1)!(n+((n+1)^2+1))=$$ $$=(n+1)!(n+n^2+2n+1+1)=$$ $$=(n+1)!(n^2+3n+2)=$$ $$=(n+1)!(n+2)(n+1)=$$ $$=(n+1)(n+2)!$$

Answer 3

Let $P(n)$ be the statement. Your inductive hypothesis in induction is to assume that $P(n)$ is true for some arbitrary value $k, k\in\mathbb{N}$. So you assume that:

$$\sum_{r=1}^k (r^2 + 1)r! = k(k+1)!$$

and now you have to try and prove that $P(k+1)$ is true if $P(k)$ is true i.e. prove that

$$\sum_{r=1}^{k+1} (r^2 + 1)r! = (k+1)(k+2)!$$

assuming that $\sum_{r=1}^k (r^2 + 1)r! = k(k+1)!$.

To progress, split the sum up as following:

$$\sum_{r=1}^{k+1} f(r) = \sum_{r=1}^{k} f(r) + f(k+1) \implies \sum_{r=1}^{k+1} (r^2 + 1)r! = \underbrace{\sum_{r=1}^{k} (r^2 + 1)r!}_{*} + \Big((k+1)^2 + 1\Big)(k+1)!$$

Substitute the equality from your inductive hypothesis for (*)

Answer 4

For n = 1 this clearly holds. Now for inductive step $n-1 \to n$: $$\sum_{r=1}^n (r^2+1)r! = \sum_{r=1}^{n-1} (r^2+1)r! + (n^2+1)n!$$ now assuming that the proposition holds for $n-1$ and substituting it for the sum we have: $$ (n-1)n! + (n^2+1)n! = n!(n^2+n+1-1) = n(n+1)n! = (n+1)!n$$

Doesn't matter if you make the step $n \to n+1$ or $n-1 \to n$.