Respuesta :
Answer: General solution is [tex] \frac{(n+1)!-1}{(n+1)!} [/tex]
Proof:
Let the equation be equal to some sum, [tex]S_n[/tex]
Now, we need to find a link between each unit.
[tex]S_n = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n}{(n+1)!}[/tex]
Let's consider a pattern by taking the first four terms.
[tex]S_1 = \frac{1}{2!} = \frac{1}{2}[/tex]
[tex]S_2 = \frac{2}{3!} = \frac{1}{3}[/tex]
[tex]S_3 = \frac{3}{4!} = \frac{1}{8}[/tex]
[tex]S_4 = \frac{4}{5!} = \frac{1}{30}[/tex]
Now, let's consider their summations and we'll call them [tex]T_n[/tex].
[tex]T_1 = S_1 = \frac{1}{2}[/tex]
[tex]T_2 = T_1 + S_2 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}[/tex]
[tex]T_3 = T_2 + S_3 = \frac{5}{6} + \frac{1}{8} = \frac{23}{24}[/tex]
[tex]T_4 = T_3 + S_4 = \frac{23}{24} + \frac{1}{30} = \frac{119}{120}[/tex]
One aspect of the sequence should be jumping out now; the numerator is always one less than the denominator.
Now, let's consider a pattern with the denominator.
We know that 2! = 2, 3! = 6, 4! = 24, and 5! = 120
We also know that they are the denominators of the sequences.
So, we can generalise our equation as:
[tex] \frac{(n+1)!-1}{(n+1)!} [/tex]
MI should be fairly simple now, if you have any problems, feel free to message me!
Proof:
Let the equation be equal to some sum, [tex]S_n[/tex]
Now, we need to find a link between each unit.
[tex]S_n = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n}{(n+1)!}[/tex]
Let's consider a pattern by taking the first four terms.
[tex]S_1 = \frac{1}{2!} = \frac{1}{2}[/tex]
[tex]S_2 = \frac{2}{3!} = \frac{1}{3}[/tex]
[tex]S_3 = \frac{3}{4!} = \frac{1}{8}[/tex]
[tex]S_4 = \frac{4}{5!} = \frac{1}{30}[/tex]
Now, let's consider their summations and we'll call them [tex]T_n[/tex].
[tex]T_1 = S_1 = \frac{1}{2}[/tex]
[tex]T_2 = T_1 + S_2 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}[/tex]
[tex]T_3 = T_2 + S_3 = \frac{5}{6} + \frac{1}{8} = \frac{23}{24}[/tex]
[tex]T_4 = T_3 + S_4 = \frac{23}{24} + \frac{1}{30} = \frac{119}{120}[/tex]
One aspect of the sequence should be jumping out now; the numerator is always one less than the denominator.
Now, let's consider a pattern with the denominator.
We know that 2! = 2, 3! = 6, 4! = 24, and 5! = 120
We also know that they are the denominators of the sequences.
So, we can generalise our equation as:
[tex] \frac{(n+1)!-1}{(n+1)!} [/tex]
MI should be fairly simple now, if you have any problems, feel free to message me!
