1+2*2!+3*3!+ ... + n*(n!)=?

2 Answers
Aug 4, 2018

(n+1)! - 1

Explanation:

sum_(r=1)^n r* r! = sum_(r=1) ^n(r+1)r! - r!

Recall : (n+1)n! = (n+1)!

=> sum_(r=1) ^n (r+1)! - r!

Method of differences:

color(red)(r = 1) : cancel(2!) color(blue)(-1!

color(red)(r = 2) : cancel(3!) - cancel(2!)
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.
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color(red)(r = n-1) : cancel(n!) - cancel((n-1)!)
color(red)( r= n) : color(blue)((n+1)!) - cancel(n!)

color(blue)(=>(n+1)! - 1

=> sum_(r=1)^n r *r! = (n+1)! - 1

Aug 4, 2018

sum_(k=1)^n k * k! = (n+1)! - 1

Explanation:

One way of solving this problem is to spot a pattern, then prove it by induction.

Looking at the first few sums, we find:

1 * 1! = 1

1 * 1! + 2 * 2! = 1+4 = 5

1 * 1! + 2 * 2! + 3 * 3! = 1+4+18 = 23

Note that we should expect a sum that involves a factorial somewhere and 23 = 24-1 = 4!-1.

Then looking at the previous values we have 5 = 6-1 = 3!-1 and 1 = 2-1 = 2!-1

So it looks like sum_(k=1)^n k * k! = (n+1)! - 1

Let's try to prove it by induction:

Let P(n) be the proposition that sum_(k=1)^n k * k! = (n+1)! - 1

Base case

Note that:

sum_(k=1)^color(blue)(1) k * k! = 1 * 1! = 1 = 2-1 = 2!-1 = (color(blue)(1)+1)! - 1

So P(1) is true.

Induction step

Suppose P(n) is true for some n.

Then:

sum_(k=1)^(color(blue)(n+1)) k * k! = sum_(k=1)^n k * k! + (n+1)(n+1)!

color(white)(sum_(k=1)^(n+1) k * k!) = (n+1)! - 1 + (n+1)(n+1)!

color(white)(sum_(k=1)^(n+1) k * k!) = (n+2)(n+1)! - 1

color(white)(sum_(k=1)^(n+1) k * k!) = ((color(blue)(n+1))+1)! - 1

So P(n+1) is true.

Conclusion

Having shown P(1) and P(n) => P(n+1) we can conclude that P(n) holds for all integers n >= 1.