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

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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.

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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.

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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.

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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.

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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.

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To prove:

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

Explanation of Solution

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Chapter C Solutions