Show that :
1! + 2 . 2! + 3 . 3! + ...... + n . n! = (n + 1)! – 1
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Given,
1!+2.2!+........+n.n!
It can be written as,
∑ r.r! , where r takes values from 1 to n
Consider r.r!
Add r! and subtract r!,
= r.r!+r!-r!
=(r+1)(r!)-r!
= (r+1)! - r!
Therefore required is ∑(r+1)! - r! , r from 1 to n.
Now expand the summation by substituting the values of r,
= (2)! - (1)! + (3)! - (2)! + (4)! - (3)! + (5)! - (4)! +..........+ (n-2)! - (n-3)! + (n-1)! - (n-2)! + (n)! - (n-1)! + (n+1)! - (n)!
You can see every term is being cancelled except (n+1)! and -(1)!,
Therefore the answer is (n+1)! - 1!
Hence 1! + 2 . 2! + 3 . 3! + ...... + n . n! = (n+1)! - 1
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