I need help to show by Induction

Question

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

$$\sum_{k=1}^n \sin(k \alpha) = \frac{\sin (\frac {n\alpha}{2}) \sin (\frac {(n+1)\alpha}{2})} {\sin (\frac {\alpha}{2})}$$

Answer 1

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

I.A

$n=1$

$\sum_{k=1}^1 k\cdot k!=1\cdot 1!=1=(1+1)!-1=2!-1=1\,\checkmark$

I.C:

For arbitrary but fixed $n\in\mathbb{N}$ let $\sum_{k=1}^n k\cdot k!=(n+1)!-1$.

I.S:

$n\mapsto n+1$

$\sum_{k=1}^{n+1} k\cdot k!=\sum_{k=1}^n k\cdot k!+(n+1)(n+1)!\stackrel{I.C}{=} (n+1)!-1+(n+1)(n+1)!$

$=(n+1)!(1+(n+1))-1=(n+1)!(n+2)-1=(n+2)!-2\,\checkmark$

A non inductive solution could be done like this:

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

Note that this sum is telescoping! So the end result is:

$=(n+1)!-1$

Answer 2

Hints:

• For the inductive step, you have to show that, if for some $n$, you have $\sum_{k=1}^n k\cdot k! = (n+1)! - 1$, then $\;\sum_{k=1}^{n+1} k\cdot k! = (n+2)! - 1$. Just write $$\sum_{k=1}^{n+1} k\cdot k!=\sum_{k=1}^{n} k\cdot k!+(n+1)\,(n+1)!$$ and make a (partial) factorisation of $(n+1)!$.
• It is faster to use that $\;\sum_{k=1}^n \sin k\alpha$ is the imaginary part of $\;\sum_{k=1}^n \mathrm e^{ik\alpha}$, which leads to a geometric progression.