Question

# Let P(k) be a statement that \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cd

Series

Let P(k) be a statement that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cdot(k+1)}=$$
for: The basis step to prove $$P(k)$$ is that at $$k = 1, ?$$ is true.
for:Show that $$P(1)$$ is true by completing the basis step proof. Left side of $$P(k)$$ and Right side of $$P(k)$$
for: Identify the inductive hypothesis used to prove $$P(k)$$.
for: Identify the inductive step used to prove $$P(k+1).$$

2021-05-19

Let the property $$P(k)$$ be $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}=\frac{k}{k+1}$$
Show that $$P(k)$$ is true for all integers $$k\geq1$$ using mathematical induction
Basis Step: $$P(k)$$ is true:
That is to show that $$\frac{1}{1\cdot2}=\frac{1}{1+1}$$
The left hand side of the equation is $$\frac{1}{1\cdot2}=\frac{1}{2}$$ and right-hand side is
$$\frac{1}{1+1}=\frac{1}{2}$$
It follows that $$\frac{1}{2}=\frac{1}{2}$$
Hence $$P(1)$$ is true.
Show that for all integers $$n\geq1$$, $$P(k)$$ is true then $$P(k+1)$$ is also true:
Suppose P(k) is true.
Then the inductive hypothesis is
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}$$
Now show that $$P(k+1)$$ is true.
That is to show that
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$$
Or, equivalently that
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$$
The left-hand side of $$P(k+1)$$ is
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{1}{k+1}(k+\frac{1}{k+2})$$
$$=\frac{1}{k+1}(\frac{k^2+2k+1}{k+2})$$
$$=\frac{1}{k+1}(\frac{(k+1)^2}{k+2})$$
$$=\frac{1}{k+1}\frac{(k+1)^2}{k+2}$$
$$=\frac{k+1}{k+2}$$
which is right hand side of $$P(k+1)$$
Hence from the principle of mathematical induction,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$$ is true, for all integers $$n\geq1$$