Let P(k) be a statement that 11⋅2+12⋅3+...+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).

Question

falhiblesw · Accepted Answer

Let the property P(k) be 11⋅2+12⋅3+13⋅4+...+1k(k+1)=kk+1 Show that P(k) is true for all integers k≥1 using mathematical induction Basis Step: P(k) is true: That is to show that 11⋅2=11+1 The left hand side of the equation is 11⋅2=12 and right-hand side is 11+1=12 It follows that 12=12 Hence P(1) is true. Show that for all integers n≥1, P(k) is true then P(k+1) is also true: Suppose P(k) is true. Then the inductive hypothesis is 11⋅2+12⋅3+13⋅4+...+1k(k+1) Now show that P(k+1) is true. That is to show that 11⋅2+12⋅3+13⋅4+...+1k(k+1)+1(k+1)(k+2)=k+1(k+1)+1 Or, equivalently that 11⋅2+12⋅3+13⋅4+...+1(k+1)(k+2)=k+1k+2 The left-hand side of P(k+1) is 11⋅2+12⋅3+13⋅4+...+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, 11⋅2+12⋅3+13⋅4+...+1n(n+1)=nn+1 is true, for all integers n≥1

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

Answered question

Answer & Explanation

New Questions in Precalculus