Prove $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers

Question

I am trying to prove by mathematical induction $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.

So far I have:

• Solved in the first case possible - $1$

• Assumed the function to be true for all positive integers

• Subbed in $k+1$ for $n$ leaving me with the following $$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$

I am unsure where to go from here? Any help would be greatly appreciated!

Answer 1

you have your base case:

Assume:

$2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$

We must show that

$2\times1!+\cdots+(n^2+1)n! + ((n+1)^2 + 1)(n+1)!=(n+1)(n+2)!$

$n(n+1)! + ((n+1)^2 + 1)(n+1)!$

By the inductive hypothesis.

$((n+1)^2+ n + 1)(n+1)!\\ (n^2 + 3n + 2)(n+1)!\\ (n+1)(n+2)(n+1)!\\ (n+1)(n+2)!$

Answer 2

The Induction Hypothesis should give you $2 \times 1! + \dots + (k^2+1)k! = k(k+1)!$. In the Induction Step, you should start from one side of the equality and try to reach the other side. Here, it makes most sense to work with the summation side, as this will allow us to easily apply the Induction Hypothesis. Applying the Induction Hypothesis, we get: $$ \begin{align*} 2 \times 1! + \dots + (k^2+1)k! + ((k+1)^2 + 1)(k+1)! &= k(k+1)! + ((k+1)^2 + 1)(k+1)! \\ &= ((k+1)^2 + k + 1) \cdot (k+1)! \\ &= (k+1 + 1)(k+1) \cdot (k+1)! \\ &= (k+1)(k+2)! \end{align*} $$

Answer 3

The key point is form the induction hypotesis

$$2\times1!+5\times2!+10\times3!+…+(k^2+1)k!=k(k+1)!$$

is that

$$2\times1!+\ldots+((k+1)^2+1)(k+1)!\stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!\stackrel{?}=(k+1)(k+2)!$$

and the latter is true indeed

$$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$