Answer to Question #261758 in Discrete Mathematics for Nathi

Question

∀n ≥ 1, Xn i=1 i(i!) = (n + 1)! − 1

Answer 1

$\displaystyle{\sum^n_{i=1} }i(i!)=(n+1)!-1$

proof by induction:

for n=1:

$\displaystyle{\sum^1_{i=1} }1\cdot(1!)=1$

$(1+1)!-1=1$

let for n=k:

$\displaystyle{\sum^k_{i=1} }i(i!)=(k+1)!-1$

then for n=k+1:

$\displaystyle{\sum^{k+1}_{i=1} }i(i!)=\displaystyle{\sum^{k}_{i=1} }i(i!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=$

$=(k+2)(k+1)!-1=(k+2)!-1$