Math, asked by saivignesh06, 10 months ago

Show that :
1! + 2 . 2! + 3 . 3! + ...... + n . n! = (n + 1)! – 1

Answers

Answered by shaikfahad3210
1

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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