1
$\begingroup$

Show that $$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$$

I got stuck after the n+1 part.

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

I'm not sure how to proceed from this part.

$\endgroup$
3
$\begingroup$

proof by Induction $$P(1):=2\cdot1!=1\cdot(1+1)!$$

Now we assume it is true for $P(n)$ then we proof it is true for $P(n+1)$, $$P(n+1):=2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! + ((n+1)^2+1)(n+1)! = (n+1)((n + 1)+1)!$$

Use the hypothesis of induction $P(n)$ and LHS of above expression reduces to $$n(n+1)!+((n+1)^2+1)(n+1)! = (n+1)!(n+n^2+2n+2) \\ =(n+1)!(n^2+3n+2)=(n+1)!(n+2)(n+1) \\ =(n+1)(n+2)! $$

Hence we get the RHS of the $P(n+1)$. So we have proved $P(n+1)$ is true whenever $P(n)$ is true.

$\endgroup$
2
$\begingroup$

You want to prove that $$\sum_{k=1}^n(k^2+1)k! = n(n+1)!.$$

First case: if $n = 1$, then $(1^2+1)1! = 2 = 1 \cdot 2 = n(n+1)!$.

Then by induction, if the formula holds for $n-1$ you find: $$\sum_{k=1}^n(k^2+1)k! = \sum_{k=1}^{n-1}(k^2+1)k! + (n^2+1)n! = (n-1)(n-1+1)! + (n^2+1)n! = (n-1)n! + (n^2+1)n! = (n+n^2)n! = n(n+1)n! = n(n+1)!$$

$\endgroup$
0
$\begingroup$

Alternatively: $$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! =\sum_{i=1}^n(i^2+1)i!=\\ \sum_{i=1}^n[(i+1)^2-i-i]i!=\\ \color{red}{\sum_{i=1}^n(i+1)\cdot (i+1)!-\sum_{i=1}^n i\cdot i!}-\color{blue}{\sum_{i=1}^n (i+1-1)\cdot i!}=\\ \color{red}{(n+1)(n+1)!-1}-\color{blue}{(n+1)!+1}= n(n + 1)!$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.