1. ## induction factorials

Prove using the principle of mathematical induction that:

1/2! +2/3! + 3/4! + ...+n/(n+1)! = 1 - 1/(n+1)! for n=1,2,3...
2(1!)+5(2!)+10(3!)+....+ (n^2+1)(n!) = n(n+1)! for n=1,2,3...

Thank you!

3. Originally Posted by christina
Prove using the principle of mathematical induction that:

1/2! +2/3! + 3/4! + ...+n/(n+1)! = 1 - 1/(n+1)! for n=1,2,3...
2(1!)+5(2!)+10(3!)+....+ (n^2+1)(n!) = n(n+1)! for n=1,2,3...

Thank you!
1

$\displaystyle \color{blue}\ \frac{1}{2!}+\frac{2}{3!}+...\frac{n}{(n+1)}!=1-\frac{1}{(n+1)!}\$ ?

If this is true, then we attempt to use this to prove

$\displaystyle \frac{1}{2!}+\frac{2}{3!}+...+\frac{n}{(n+1)!}+\fr ac{(n+1)}{([n+1]+1)!}=1-\frac{1}{([n+1]+1)!}$

Proof

$\displaystyle 1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\left(\frac{1}{(n+1)!}-\frac{n+1}{(n+2)!}\right)=1-\left(\frac{n+2}{(n+2)!}-\frac{n+1}{(n+2)!}\right)$$\displaystyle =1-\frac{n-n+2-1}{(n+2)!}=1-\frac{1}{(n+2)!} Now test an initial term 2 \displaystyle \color{blue}\ 2(1!)+5(2!)+10(3!)+....+(n^2+1)(n!)=n(n+1)! ? If this is true, then we try to use this to prove if \displaystyle 2(1!)+5(2!)+.....+(n^2+1)(n!)+\left((n+1)^2+1\righ t)(n+1)!=(n+1)(n+2)! Proof \displaystyle n(n+1)!+\left((n+1)^2+1\right)(n+1)!=n(n+1)!+\left (n^2+2n+1+1\right)(n+1)! \displaystyle =(n+1)!\left(n+n^2+2n+2\right)=(n+1)!\left(n^2+3n+ 2\right)$$\displaystyle =(n+1)!(n+2)(n+1)=(n+2)!(n+1)$

Now test for an initial term