To prove: The formula for all natural number using the principle of induction.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

18. Using the method of variation of parameter, a particular solution to y′′ + 16y = 4 sec(4t) isyp(t) = u1(t) cos(4t) + u2(t) sin(4t). Then u2(t) is equal toA. 1 B. t C. ln | sin 4t| D. ln | cos 4t| E. sec(4t)

Question 4. Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1, 3). You don't have an equation for S but you know that the curves r1(t) = (2 + 3t, 1 — t², 3 − 4t + t²) r2(u) = (1 + u², 2u³ − 1, 2u + 1) both lie on S. (a) Check that both r₁ and r2 pass through the point P. 1 (b) Give the expression of the 074 in two ways Ət ⚫ in terms of 32 and 33 using the chain rule მყ ⚫ in terms of t using the expression of z(t) in the curve r1 (c) Similarly, give the expression of the 22 in two ways Əz ди ⚫ in terms of oz and oz using the chain rule Əz მყ • in terms of u using the expression of z(u) in the curve r2 (d) Deduce the partial derivative 32 and 33 at the point P and the equation of მე მყ the tangent plane

Coast Guard Patrol Search Mission The pilot of a Coast Guard patrol aircraft on a search mission had just spotted a disabled fishing trawler and decided to go in for a closer look. Flying in a straight line at a constant altitude of 1000 ft and at a steady speed of 256 ft/s, the aircraft passed directly over the trawler. How fast (in ft/s) was the aircraft receding from the trawler when it was 1400 ft from the trawler? (Round your answer to one decimal places.) 1000 ft 180 × ft/s Need Help? Read It SUBMIT ANSWER

Answer 2

Question

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Answer 6

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Answer 7

Chapter C, Problem 1E

To determine

To prove:

The formula for all natural number using the principle of induction.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given info.

$1 + 2 + 3 + .............. + n = \frac{n (n + 1)}{2}$

If $(1)$ the statement is true for $n = 1$ and

$(2)$ When a statement is true for a natural number $n = k$ then it will also be true for its successor $n = k + 1$ .

Then statement will be true for all natural number n. this is called principle of mathematical induction.

Proof:

Show that $P (1)$ is true

Left hand side $(L . H . S .)$

$P (1) = 1$

R.H.S.

$\begin{matrix} P (1) = \frac{1 \times (1 + 1)}{2} \\ = \frac{1 \times 2}{2} \\ = 1 \end{matrix}$

$L . H . S . = R . H . S .$

Thus, the statement is true for $n = 1$

Now, follow induction step.

If $P (n)$ is true for $n = k$ then it will also be true for $n = k + 1$

Let us assume that $P (k)$ is true condition that means

$P (k) = 1 + 2 + 3 + ......... + k = \frac{k (k + 1)}{2}$

Then by taking above statement true, proceed and prove that its successor will also be true.

$P (k + 1) = 1 + 2 + 3 + ........... + k + (k + 1) = \frac{(k + 1) (k + 2)}{2}$

L.H.S.

$1 + 2 + 3 + .......... + k + k + 1$

As we know that

$(1 + 2 + 3 + ........... + k) = \frac{k (k + 1)}{2}$

Substitute the value

$\begin{matrix} P (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{k (k + 1) + 2 (k + 1)}{2} \end{matrix}$

Take $(k + 1)$ as common factor

$P (k + 1) = \frac{(k + 1) (k + 2)}{2}$

This is the value of R.H.S.

Now, we have full-filled both conditions of the principle of mathematical induction. The formula is therefore true for every natural number.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. C - Prob. 1E Ch. C - Prob. 2E Ch. C - Prob. 3E Ch. C - Prob. 4E Ch. C - Prob. 5E Ch. C - Prob. 6E Ch. C - Prob. 7E Ch. C - Prob. 8E Ch. C - Prob. 9E Ch. C - Prob. 10E

To prove: The formula for all natural number using the principle of induction.

Concept explainers

Videos

To prove:

The formula for all natural number using the principle of induction.

Explanation of Solution

Want to see more full solutions like this?

Chapter C Solutions