Calculus: Early Transcendentals
Calculus: Early Transcendentals
4th Edition
ISBN: 9781319050740
Author: Jon Rogawski, Colin Adams, Robert Franzosa
Publisher: W. H. Freeman
Question
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Chapter C, Problem 1E
To determine

To prove:

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

Expert Solution & Answer
Check Mark

Explanation of Solution

Given info.

1+2+3+..............+n=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.

P(1)=1×(1+1)2=1×22=1

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=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)=(k+1)(k+2)2

L.H.S.

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

As we know that

(1+2+3+...........+k)=k(k+1)2

Substitute the value

P(k+1)=k(k+1)2+(k+1)=k(k+1)+2(k+1)2

Take (k+1) as common factor

P(k+1)=(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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