Question

Compute the sum: S=1!.1+2!.2+3!.3+....+2011!.2011.

Answer 1

(2012)!-1

Answer 2

\[S=1!\cdot1+2!\cdot2+3!\cdot3+....+n!\cdot n=(n+1)!-1\]

Answer 3

yes - if we set the nth term as:\[n*n!\]then the term before this is:\[n!-(n-1)!\]so I believe a lot of the terms cancel out leaving you with the @Zarkons answer

Answer 4

which comes from expansion of:\[(n-1)!*(n-1)\]

Answer 5

http://openstudy.com/#/updates/4ececc98e4b04e045af1b27a esay question

Answer 6

it is a simple proof by induction

Answer 7

U GUYS PLEASE HELP ME

Answer 8

I thought for induction you had to be given an equation upfront and then prove its true?

Answer 9

I only know it of hand because I just assigned it this semester to my discrete math class

Answer 10

and I did give you an equation ;)

Answer 11

where? I am not in any maths class - I just do maths as a hobby

Answer 12

this is one of those problems where I would have looked at smaller numbers to see if i could find a pattern

Answer 13

or are you saying that the original question was set as prove ...=(n+1)!-1

Answer 14

yes

Answer 15

ah - ok - then induction would work - thanks

Answer 16

but looking at the terms, I believe they can be written out in reverse order (i.e. starting from the nth term) as:
(n*n!) + (n!-(n-1)!) + ((n-1)!-(n-2)!) + ...

Answer 17

so all the intermediate factorials cancel out

Answer 18

yes

Answer 19

leaving:
n*n! + n! - 1 = n!(n+1) - 1 = (n+1)! - 1

Answer 20

ok - so pattern recognition works too :-)

Answer 21

almost always more than one way to do things

Answer 22

does pattern recognition count as a proper proof in maths?

Answer 23

usually....but I asked for a proof by induction in my particular class

Answer 24

I was just curious as I am interested in mathematics and was wondering how "strict" proofs have to be before they are considered "worthy"

Answer 25

doing it your way is find....I covered induction and I wanted to make sure that they new how to do it using indiction

Answer 26

*fine

Answer 27

ok - thanks again for taking the time to respond.

Answer 28

no problem :)

Answer 29

You could make the sum into a telescoping sum too:\[\sum_{k=1}^{2011}k\cdot k!=\sum_{k=1}^{2011}((k+1)-1)\cdot k!=\sum_{k=1}^{2011}(k+1)!-k!=2012!-1!\]

Answer 30

yes...that is what asnaseer did.

Answer 31

ah my bad, i didnt take the time to read it all >.<

Answer 32

although I ever realised its called a telescoping sum :-)

Answer 33

**never

Answer 34

no probs @joemath - your posting has just incremented my maths vocabulary :-)

Answer 35

zarkon you made a correction to one word in that sentence
you should had made two corrections
new should have been knew
and you found the one with find which was suppose to be fine
i just thought i would correct you on something since it makes me feel good lol