Question

a) For which positive integers n is \(11n + 17 \le {2^n}\)?

b) Prove the conjecture you made in part (a) using mathematical induction.

Short answer

a) Thus, \(11n + 17 \le {2^n}\) is true.

b) The given statement is true\(n \ge 7\).

Step-by-step solution

Principle of Mathematical Induction.

We know the statement of mathematical induction.

If the statement is true for

\(p\left( 1 \right)\)is true.

if we assume\(n = k\)is true then\(p\left( k \right)\)is true.

if we assume \(n = k + 1\) is true then all the value of n is true in the statement.

(a) here we solve which positive integer n is \(11n + 17 \le {2^n}\)?

1) Explanation.

In the given question n is a positive integer and also,

Given , \(11n + 17 \le {2^n}\_\_\_\_(i)\).

put\(n = 1,2,3,4........\). In equation \(\left( i \right)\)

The given expression is\(11n + 17 \le {2^n}\).

2) solution

Given in the expression\(11n + 17 \le {2^n}\).

Case:1

If put n =1 in the given expression, then we get.

\(\begin{aligned}{c} = 11 \times 1 + 17 \le {2^1}\\ = 11 + 17 \le 2\\ = 28 \le 2\end{aligned}\)

Here LHS = 28 and RHS = 2.

LHS > RHS

This is not satisfactory.

Case 2 : Put\(n = 2\).

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 2 + 17 \le {2^2}\\ = 22 + 17 \le 4\\ = 39 \le 4\end{aligned}\)

LHS =39 and RHS = 4

LHS > RHS

This is not satisfactory.

Case 3: Put \(n = 3\)

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 3 + 17 \le {2^3}\\ = 33 + 17 \le 8\\ = 50 \le 8\end{aligned}\)

LHS = 50 and RHS = 8

LHS >RHS

This is not satisfactory.

Case:-4

\(n = 4\)

\(11n + 17 \le {2^n}\)

\( = 11 \times 4 + 17 \le {2^4}\)

\( = 61 \le 16\)

LHS = 61 and RHS = 16

LHS > RHS

This is not satisfactory.

Case 5: \(n = 5\)

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 5 + 17 \le {2^5}\\ = 55 + 17 \le 32\\ = 72 \le 32\end{aligned}\)

LHS = 72 and RHS = 32

LHS > RHS

This is not satisfactory.

Case 6: \(n = 6\)

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 6 + 17 \le {2^6}\\ = 66 + 17 \le 64\\ = 83 \le 64\end{aligned}\)

LHS = 83 and RHS = 64

LHS > RHS

This is not satisfactory.

Case 7: \(n = 7\)

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 7 + 17 \le {2^7}\\ = 77 + 17 \le 128\\ = 94 \le 128\end{aligned}\)

LHS = 94 and RHS = 128

LHS < RHS

This is satisfied.

Now we have finally found the positive integer\(n = 7\).

Hence the given condition \(11n + 17 \le {2^n}\) is true.

  Step 3: (b)  the conjecture you made in part (a) using mathematical induction.

1) Explanation.

In the given (a) part\(11n + 17 \le {2^n}\).

We have to solve this expression using mathematical induction.

2) Solution.

If put \(n = 7\) in the given statement.

Then we get,

\(\begin{aligned}{c}11n + 17 \le {2^n}\\ = 11 \times 7 + 17 \le {2^7}\\ = 77 + 17 \le 128\\ = 94 \le 128\end{aligned}\)

Hence LHS < RHS .

So, \(n = 7\) implies that \(p\left( 7 \right)\) is true.

Now, if we put \(n = k\)in the given statement\(p\left( k \right)\).

Then we get

\(11 \times k + 17 \le {2^k}\)

Here the given statement is true.

Here now but \(n = k + 1\) then,

\(\begin{aligned}{c} = 11\left( {k + 1} \right) + 17\\ = 11k + 11 + 17\\ = \left( {11k + 11} \right) + 17\end{aligned}\)

Here solve,

\(\begin{aligned}{l}11\left( {k + 1} \right) = {2^k}\\ \Rightarrow 11\left( {k + 1} \right) + 17 < {2^k} + 11\\ \Rightarrow 11\left( {k + 1} \right) + 17 < {2^{k + 1}}\end{aligned}\)

Hence, \(p\left( {k + 1} \right)\) is true.

Hence, the given statement is true\(n \ge 7\). this is the principle of mathematical induction.