# Thread: Ugh, another inductive proof

1. ## Ugh, another inductive proof

I have to prove this inductively, bare with me if the lingo isn't perfect.

Prove that: 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1

Proof:
Base Case: n=1;
LHS = 1(1!) = 1
RHS = (1+1)! - 1 = 2! - 1 = 1
Done;

Now, assume true for n; Prove true for n=n+1.
Let n= n+1
RHS = ( (n+1)+1)! - 1 = (n + 2)! - 1
LHS = 1(1!) + 2(2!) + ... + n(n!) + (n+1)((n+1)!)

now, since we stated that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1 we can sub that in its place such that the equation looks like the following:

LHS = (n+1)! - 1 + (n+1)((n+1)!)

Now here is where I am stuck ... I have tried factoring and combining like terms in every way I can think of ... anyone want to lend a hand =D
TY

2. Originally Posted by mcox05
Prove this inductively 1(1!) + 2(2!) + ... + n(n!) = (n+1)! -1
You have the base case.
Suppose that this is true: $\sum\limits_{k = 1}^N {k\left( {k!} \right)} = \left( {N + 1} \right)! - 1$.
Note that: $\left( {N + 2} \right)! - 1 = \left( {N + 1} \right)!\left[ {N + 2} \right] - 1 = \left( {N + 1} \right)!\left[ {1 + \left( {N + 1} \right)} \right] - 1$
$\begin{array}{lcl}
{\sum\limits_{k = 1}^{N + 1} {k\left( {k!} \right)} } & = & {\sum\limits_{k = 1}^N {k\left( {k!} \right)} + \left( {N + 1} \right)\left[ {\left( {N + 1} \right)!} \right]} \\
{} & = & {\left[ {\left( {N + 1} \right)! - 1} \right] + \left( {N + 1} \right)\left[ {\left( {N + 1} \right)!} \right]} \\
{} & = & {\left( {N + 2} \right)! - 1} \\
\end{array}$