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

Kaycee Roche

Kaycee Roche

Answered question

2021-10-28

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

funblogC

funblogC

Skilled2021-10-29Added 91 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)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?