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

Nannie Mack

Nannie Mack

Answered question

2021-05-18

Let P(k) be a statement that 112+123+...+1k(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).

Answer & Explanation

falhiblesw

falhiblesw

Skilled2021-05-19Added 97 answers

Let the property P(k) be 112+123+134+...+1k(k+1)=kk+1
Show that P(k) is true for all integers k1 using mathematical induction
Basis Step: P(k) is true:
That is to show that 112=11+1
The left hand side of the equation is 112=12 and right-hand side is
11+1=12
It follows that 12=12
Hence P(1) is true.
Show that for all integers n1, P(k) is true then P(k+1) is also true:
Suppose P(k) is true.
Then the inductive hypothesis is
112+123+134+...+1k(k+1)
Now show that P(k+1) is true.
That is to show that
112+123+134+...+1k(k+1)+1(k+1)(k+2)=k+1(k+1)+1
Or, equivalently that
112+123+134+...+1(k+1)(k+2)=k+1k+2
The left-hand side of P(k+1) is
112+123+134+...+1k(k+1)+1(k+1)(k+2)
=kk+1+1(k+1)(k+2)
=1k+1(k+1k+2)
=1k+1(k2+2k+1k+2)
=1k+1((k+1)2k+2)
=1k+1(k+1)2k+2
=k+1k+2
which is right hand side of P(k+1)
Hence from the principle of mathematical induction,
112+123+134+...+1n(n+1)=nn+1 is true, for all integers n1

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