Help proof by induction

Question

I just got my first induction assignment in a new course. They want me to prove by induction that: $$\sum_{s=1}^k s*s! = (K+1)!-1.$$

The way I understand induction is that I test for the first value. Then for n and then n+1, to show the given expression is true. I've done it for 1 and n, however I'm stuck at n+1. $$\sum_{s=1}^{n+1} s*s!=(n+1)!-1+(n+1)*(n+1)!$$ Using maple I can see the expression is $-1+(n+2)!$ (which is true) however i dont know how to reduce/rewrite $(n+1)!-1+(n+1)(n+1)!$ to $-1+(n+2)!.$$

I've asked my friends and a older student, but to no avail. I'm hoping you guys can help.

Welcome to MSE. Please type your questions instead of posting images. Images can't be browsed and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial — saulspatz, Feb 18 '21 at 14:44
As for your specific troubles... recall that $ac+bc = (a+b)c$. This is true regardless how complicated or simple of expressions $a,b,c$ happen to be. Here, include an extra invisible factor of $1$ on the first term (which is always allowed as it doesn't change anything) to get $\color{red}{1}\cdot\color{blue}{(x+1)!} + \color{green}{(x+1)}\cdot\color{blue}{(x+1)!} = (\color{red}{1}+\color{green}{(x+1)})\cdot\color{blue}{(x+1)!}$. Next, simplify $1+x+1$ as $x+2$, and then simplify $(x+2)\cdot (x+1)!$ as $(x+2)!$ using basic properties of factorials. — JMoravitz, Feb 18 '21 at 14:49
As for generic tips regarding induction, I recommend checking out How to write a clear induction proof as well as What makes induction a valid proof technique. — JMoravitz, Feb 18 '21 at 14:51
Does this answer your question? What is the telescoping series? $\sum_{k=1}^{n} k \cdot k!$ — , Feb 18 '21 at 15:03
@JitendraSingh Note, the OP is asking for proof by induction. — VIVID, Feb 18 '21 at 15:13

score 1 · Accepted Answer · answered Feb 18 '21 at 15:01

Just factor out the common factor $(n+1)!$ $$\begin{align} \color{blue}{(n+1)!}-1+(n+1)\cdot\color{blue}{(n+1)!} &= \color{blue}{(n+1)!}(1+(n+1))-1 \\ &= \color{blue}{(n+1)!}(n+2) - 1 \\ &= (n+2)!-1\end{align}$$

For the sake of another direct proof, see this answer.

score 0 · Answer 2 · answered Feb 18 '21 at 15:23

First proof the statement for some value, which you did (induction basis).

Then induction step: Let's assume that the statement you want to proof holds for K = n. Then you have to establish that it also holds for n+1.

So we know that $\sum_{s=1}^{n} s \cdot s! = (n+1)! - 1$ (induction hypothesis)

$\sum_{s=1}^{n+1} s \cdot s! = \sum_{s=1}^{n} s \cdot s! + (n+1) \cdot (n+1)! = (n+1)! - 1 + (n+1) \cdot (n+1)! = (n+1)! + (n+1) \cdot (n+1)! - 1 = (1 + (n+1)) \cdot (n+1)! - 1 = (n+2) \cdot (n+1)! - 1 = (n+2)! - 1 = ((n+1)+1)! - 1$

So then we know that the statement also holds for n+1. This concludes the proof.

So in proofs by induction you explicitly use the induction hypothesis for the induction step.

Help proof by induction

2 Answers2