Question

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

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asked 2021-05-18

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 \(P(k+1).\)

Answers (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 \(n\geq1\)

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