1(1!) + 2(2!) + 3(3!) + ... + 49(49!) + 50(50!) = ?

[Math Tutor]

Calculate the sum:
1(1!) + 2(2!) + 3(3!) + ... + 49(49!) + 50(50!) = ?

[Guest]

Spoiler: show

[tex]n(n!) = (n+1)!-n![/tex]
using this identity the expression is clearly a telescoping sum meaning the answer is 51!-1! = 1551118753287382280224243016469303211063259720016986111999999999999

R. Baber.

[Guest]

Not my kind of sums but just curious and using my antique Casio fx-82B calculator

50! = 3.0414093 x 10^64

50 x 50! = 1.5207047 x 10^66 ......xxxxxx

51! = 1.5511188 x 10^66

51! - 50! = 1.5207047 x 10^66 ......xxxxxx

this is different from R Baber's answer and I only considered 1 term...?????
but it checks the identity given works.

[Guest]

Just in case my previous post caused any confusion:

Spoiler: show

[tex]n(n!) = ((n+1)-1)(n!) = (n+1)(n!)-1(n!) = (n+1)!-n![/tex]
Substituting this into the expression
[tex]50(50!) + 49(49!) + ... + 2(2!) + 1(1!)[/tex]
gives
[tex](51! - 50!) + (50! - 49!) + ... + (3! - 2!) + (2! - 1!)[/tex]
When the brackets are removed and the expression is simplified most of the terms cancel out leaving
[tex]51!-1![/tex]

R. Baber.

[Guest]

The bit I don't understand...
Your first line states....
n(n!)=(n+1)!−n!

Which I attempted to calculate as follows....using 50! and 51!

50! = 3.0414093 x 10^64

50 x 50! = 1.5207047 x 10^66 ......xxxxxx

51! = 1.5511188 x 10^66

51! - 50! = 1.5207047 x 10^66 ......xxxxxx

And showed that both sides of the identity are equal...OK

Your last line states it is equal to..... 51! -1!...for what I assume you took as 49! and 50!

49! = 6.0828186 x 10^62

49 x 49! = 2.9805811 x 10^64 ....======

50! = 3.0414093 x 10^64

50! - 49! = 2.9805811 x 10^64.....======

which is not same as 51! - 1! .....OR is it...????.

[Guest]

Sorry....I think the penny has dropped.....

you are converting each term in the series.....

eg.... 50x50! to (51! - 50!)....etc
then the + and - cancel.....
leaving (51! - 1!) = 1.5511188 x 10^66
This is the sum of all the terms... from 1 to 50...is that correct...???

[Guest]

You got it.

R. Baber.

[Guest]

Thanks