#1+2*2!+3*3!+ ... + n*(n!)=#?
2 Answers
Explanation:
Recall :
Method of differences:
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
Then looking at the previous values we have
So it looks like
Let's try to prove it by induction:
Let
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
Induction step
Suppose
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
Conclusion
Having shown