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

Question

I'm to prove this by mathematical induction: Edited: I made a typo error. $\sum_{k=1} ^n (k^2+1)k!=n(n+1)!$

I made the test and the rightside is true.

So I tested: $N+1$

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

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

$N!\left[N(N+1)+N^2+1\right]$

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

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

Is this a valid proof?

Answer 1

$\sum _{k=1} ^{N+1} (k^2+1)k!$=$[(N+1)^2 +1]!+N(N+1)!\\ =(N+1)![(N^2+3N+2]\\ =(N+1)(N+1)(N+2)\\ =(N+1)(N+2)!$

Answer 2

Your inductive step doesn't work because your expression doesn't simplify to $(N+2)!(N+1)$.

Note that $$\sum_{k=1}^n(k^2+1)k!=\sum_{k=1}^n((k+2)!-3(k+1)!+2\cdot k!)\\=\sum_{k=1}^n((k+2)!-(k+1)!+2\cdot k!-2\cdot (k+1)!)\\=(n+2)!-2\cdot (n+1)!=(n+1)!n.$$

Answer 3

There are various problems, including

• the thing you are trying to prove is not correct, for example when $n=2$
• you did not end up showing the inductive hypothesis was true for $N+1$
• you seem to have added $(N^2+1)N!$ rather than $(N+1+1)\,(N+1)!$

So let's illustrate proving something that is true: $$\sum\limits_{k=1} ^n k\cdot k!=(n+1)!-1$$

First we check that it is correct at the start, when $n=1$: we have $1 \cdot 1!=1$ on the left hand side and $2!-1=1$ on the right and these are indeed equal

Then we assume it is true for $n=m$ and consider what happens trying to go forward a step:

$\sum\limits_{k=1} ^{m+1} k\cdot k! = \sum\limits_{k=1} ^m k\cdot k! + (m+1)\cdot (m+1)! \\ =(m+1)!-1+(m+1)\cdot (m+1)! \\= (m+1)!(1+m+1)-1 \\= (m+2)! -1$

which shows it would also be true for $n=m+1$

and therefore, by induction, it is true for all positive integer $n$