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a sequence $a_1,a_2,a_3,...$ is defined by letting $a_1 = 2$ and $a_k = \frac{a_k-1}{k}$ for all integers $k\ge2$.

use mathematical induction to prove $a_n = \frac{2}{n!}$ for all integers $n\ge1$ is a general form of the sequence.

let $p(1)$ be the statement $a_1 = \frac{2}{1!} = 2$

now the left hand side of $p(1)$ is 2. the right hand side is also 2 so the statement holds.

now assume $p(k)$ holds for $k\ge1$ so assume that $a_k = \frac{2}{k!}$

the statement $p(k+1)$ states: $a_k+1 = \frac{2}{(k+1)!}$

the left hand side of $$p(k+1) = a_k+1$$ $$ = \frac{a_k}{k+1}$$ $$ = \frac{2}{k!(k+1)}$$ $$ = \frac{2}{k!(k+1)}$$ $$ = \frac{2}{(k+1)!}$$ which equals right hand side of $p(k+1)$

can someone please explain to me how $\frac{2}{k!(k+1)}$ becomes $\frac{2}{(k+1)!}$?

I thought that $\frac{2}{k!(k+1)}$ would become $\frac{2}{k^2!+k!}$

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It's just the definition of the factorial: $$(k+1)!=1\cdot2\cdot...\cdot k\cdot (k+1)=k!(k+1)$$

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  • $\begingroup$ that is so simple, i never thought of that, thanks! $\endgroup$ – kr1s Oct 6 '17 at 3:20
  • $\begingroup$ @kr1s You are welcome! $\endgroup$ – Michael Rozenberg Oct 6 '17 at 3:21

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