proof by induction - creating summations?

Question

I have two proofs I need to do that I can not figure out how to turn into summations in order to solve.

I have to write it like this $$\sum_{i=1}^{n}i=n$$

#1 is related to http://math.stackexchange.com/questions/697892/proof-that-23n-1-is-divisible-by-11-for-all-positive-integers-n — lab bhattacharjee, Mar 05 '14 at 06:34
i see the connection but I still don't see how to write it out as a $\sum_{k=1}^{n}i=n$ style summation — Mikky Davey, Mar 05 '14 at 06:39

score 1 · Answer 1 · answered Mar 05 '14 at 06:35

1

HINT:

$#2$

If $\displaystyle n^n\ge n!,$

$\displaystyle (n+1)^{n+1}=(n+1)\cdot (n+1)^n>(n+1)n^n\ge (n+1)\cdot n!$

answered Mar 05 '14 at 06:35

lab bhattacharjee

score 0 · Answer 2 · answered Mar 05 '14 at 06:37

0

$4^n = (3+1)^n$. When you develop the product, there are only multiple of 3 and the $+1$ obtained multiplying 1 by itself $n$ times. So $4^n - 1$ is a multiple of 3.
$n! = n(n-1)\cdots 2\times 1$, and there are $n$ factors. They are all $\le n$, so $$ n! \le n^n $$

answered Mar 05 '14 at 06:37

mookid

do you know how i could write that as a summation? – Mikky Davey Mar 05 '14 at 06:46

2 Answers2