Challenge Problem #2

Soroban · **Posted:** Tue, 12 Aug 2003 13:18:59 UTC

$2)\;$ Find the sum of this series:

. $_{n}$
$\sum\ k(k!)\; =\; 1(1!) + 2(2!) + 3(3!) + ... + n(n!)$
$^{k=1}$

bugzpodder · **Joined:** Sun, 4 May 2003 16:04:19 UTC **Posts:** 2906

yay!

1+1*1!+2*2!+...+n*n!
notice k*k!+k=(k+1)*k!=(k+1)!

so if i add one, i get:
(n+1)!
so the answer is:
(n+1)!-1

Soroban · **Posted:** Tue, 12 Aug 2003 16:54:01 UTC

Well done, Bugz!

For those of you who didn't follow Bugz' elegant solution,
I'll post a detailed explanation in a day or so.

Soroban · **Posted:** Tue, 12 Aug 2003 20:41:22 UTC

Here's the book's solution to Challnge Problem #2.

If you're still working on it, don't look beyond the wavy line.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We have: S = 1(1!) + 2(2!)+ 3(3!) + 4(4!) + ... + n(n!)

Add the quantity 1! + 2! + 3! + ... + n!

and subtract it.

$S = 1(1!) + {\bf1!} + 2(2!) + {\bf 2!} + 3(3!) + {\bf 3!} + ... + n(n!) + {\bf n!}$
. . . $-\ ({\bf 1! + 2! + 3! + ... + n!})$

= 2(1!) + 3(2!) + 4(3!) + ... + (n+1)(n!) - (1! + 2! + 3! + ... + n!)

= 2! + 3! + 4! + ... + (n + 1)! - (1! + 2! + 3! + ... + n!)

Every cancels out except: $-1!\ and\ (n + 1)!$

Answer: S = (n + 1)! - 1

SilverSprite · **Posted:** Fri, 15 Aug 2003 06:06:47 UTC

This reminds me of an arml question.

S.O.S. Mathematics CyberBoard

Challenge Problem #2

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