Determine $n$ in equality: $1+2(2!)+3(3!)+…+n(n!)=719$.(Answer:5)
I tried to separate the terms but I didn't succeed
$ 1+2(2!)+3.3(2!)+4.4.3(2!)+\ldots+n.n((n-1)!)=719\\ 2!(2+3^2+4^2.3+5^2.4.3\ldots)+n^2((n-1)!)=718$
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1See what happens when you write each summand $kk!$ as $[(k+1)-1]k!=(k+1)!-k!$. – plop Jan 06 '23 at 22:50
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@AnneBauval I don't understand how he arrived at this equality: $ \sum_{n = 1}^{N} n\cdot n! = (1 + N)! - 1$ – peta arantes Jan 07 '23 at 10:39
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Just sum up the equality of the accepted answer (which is the same as owl's hint above): it gives you a telescoping sum. – Anne Bauval Jan 07 '23 at 10:42
2 Answers
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$$ \begin{aligned} \sum_{k = 1}^n{k\cdot k!} &= \sum_{k = 1}^n{\left[(k+1) - 1\right]\cdot k!} = \sum_{k = 1}^n{\left[(k+1)\cdot k! - k!\right]} = \sum_{k = 1}^n{\left[(k+1)! - k!\right]} = \\ &= \left[2!-1!\right] + \left[3!-2!\right] + \ldots + \left[(n-1)!-(n-2)!\right] + \left[n!-(n-1)!\right] + \left[(n+1)!-n!\right] = \\ &= \left[\color{blue}{2!}-1!\right] + \left[\color{red}{3!}-\color{blue}{2!}\right] + \ldots + \left[\color{blue}{(n-1)!}-\color{red}{(n-2)!}\right] + \left[\color{green}{n!}-\color{blue}{(n-1)!}\right] + \left[(n+1)!-\color{green}{n!}\right] = \\ &= \left[\text{all the colored terms are canceled out}\right] = \\ &= (n+1)!-1 = 719 \Leftrightarrow (n+1)! = 720 \Rightarrow n = \ldots \end{aligned} $$
Eugene
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There are two ways, the easy way and the hard way.
Obviously 719 = 6!-1. 5 * 5! is a bit smaller than 6!. Your guess is that the sum up to n(n!) equals (n+1)!-1, you prove that by induction, then you have (n+1)! = 720, n+1 = 6, n = 5.
Or you just add 1, $2*2!=4$, $3*3!=18$, $4*4!=96$, $5*5!$ = 600, the total is 719, so n= 5.
People will disagree which is the easy and which is the hard way.
gnasher729
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