Recitation 1 and 2 MA114:003–005 14 January 1999 Spring 1999

First, let us explain the use of the ∑ for summation. The notation

∑_{k=1}^{n} f(k)

means to evaluate the function f(k) at k = 1, 2, …, n and add up the results. In other words:

∑_{k=1}^{n} f(k) = f(1) + f(2) + … + f(n).

For example:

∑_{k=1}^{4} k^{2} = 1 + 4 + 9 + 16,

∑_{k=1}^{n} 2k − 1 = 1 + 3 + 5 + … + 2n − 1,

and

∑_{k=1}^{n} 1 = n.

The text only briefly mentions mathematical induction in the section of problem solving pp. 87–93. The main ideas of mathematical induction and a few examples appear below.

The principle of mathematical induction is used to establish the truth of a sequence of statements or formula which depend on a natural number, n = 1, 2, . . .. The principle is:

Principle of mathematical induction Suppose that P_{n} is a sequence of statements depending on a natural number n = 1, 2, . . .. If we can show that:

• P_{1} is true

• For N = 1, 2, …: If P_{N} is true, then P_{N+1} is true.

Then, we can conclude that all the statements P_{n} are true.

The reason this works is that if we know P_{1} is true, then the second step allows us to conclude P_{2} is true. Now that we know P_{2} is true, then the second step allows us to conclude P_{3} is true. If we repeat this n − 1 times, we know that P_{n} is true. This principle is useful because it allows us to prove an infinite number of statements are true in just two easy steps! Below are several examples to illustrate how to use this principle. In the examples below, you should note that, in the second step, the key point is to show how to go from the statement P_{N} to P_{N+1}. See pages 88 and 91 of the text for additional discussion of mathematical inductions.

Example Show that for n = 1, 2, 3, …, the number 4^{n} − 1 is a multiple of 3. Solution Step 1. We need to show this is true when n = 1. This is easy since 4^{1} − 1 = 4 − 1 = 3 and 3 is divisible by 3. Step 2. We suppose that 4^{N} − 1 is a multiple of 3 and we want to use this assumption to show that 4^{N+1} − 1 is a multiple of 3. Our assumption for N means that for some whole number M, 4^{N} − 1 = 3M. Now 4^{N+1} − 1. If we add and subtract 4, we have

4^{N+1} − 1 = 4^{N+1} − 4 + 4 − 1 = 4(4^{N} − 1) + 3.

Now we use our assumption that 4^{N} − 1 is a multiple of 3 to replace 4^{N} − 1 by 3M and obtain that

4^{N+1} − 1 = 4 · 3M + 3 = 3(4M + 1).

Thus we have shown that 4^{N+1} − 1 is a multiple of 3.

Example Show that for n = 1, 2, …, we have

∑_{j=1}^{n} 2j = n(n + 1).

Solution Step 1. If n = 1, then n(n + 1) = 1 · 2 = 2. Also,

∑_{j=1}^{1} 2j = 2.

Thus both sides are equal if n = 1. Step 2. Now suppose that the formula is true for N and consider the sum

∑_{j=1}^{N+1} 2j = ∑_{j=1}^{N} 2j + 2(N + 1).

We use our assumption that ∑_{j=1}^{N} 2j = N(N + 1) to conclude that

∑_{j=1}^{N+1} 2j = N(N + 1) + 2(N + 2).

Simplifying this last expression gives

N(N + 1) + 2(N + 1) = N^{2} + N + 2N + 2 = N^{2} + 3N + 2 = (N + 2)(N + 1).

Since (N + 2)(N + 1) = (N + 1 + 1)(N + 1), we have shown that the formula

∑_{j=1}^{N+1} 2j = (N + 1 + 1)(N + 1)

is true. This completes the proof by induction.

Below is a selection of problems related to mathematical induction. These problems will not be collected or graded. However, you should understand how to work each of these problems. You should begin working on these problems in groups in recitation. You will probably want to finish these problems outside of class. If you have questions, please ask your TA or instructor.

1. Use mathematical induction to prove that for each n = 1, 2, …,

∑_{k=1}^{n} k = n(n + 1)/2.

2. (a) Find a simple formula for

∑_{k=1}^{n}(k + 1)^{2} − k^{2} = 2^{2} − 1 + (3^{2} − 2^{2}) + … + n^{2} − (n − 1)^{2} + (n + 1)^{2} − n^{2}.

(b) Using your answer to part a), find another proof of the formula in the previous problem. To do this you should simplify each summand on the left.

3. Use mathematical induction to prove that

∑_{k=1}^{n} k^{2} = n(n + 1)(2n + 1)/6.

4. Use mathematical induction to prove that

∑_{j=1}^{n} j^{3} = [\frac{n(n + 1)}{2}]^{2} .

5. Let f_{1}(x) = x − 2 and then define f_{n} for n = 1, 2, … by f_{n+1}(x) = f_{1} ∘ f_{n}(x). (It is the principle of mathematical induction which tells us that these two statements suffice to define f_{n} for all n!) Use mathematical induction to prove that

f_{n}(x) = x − 2n.

6. Let f(x) = sin(2x). Prove that for n = 1, 2, …,

\frac{d^{2n}}{dx^{2n}} f(x) = (−4)^{2n} sin(2x).

7. Let f(x) = xe^{x}. Compute f′, f′′, and f′′′. Guess a formula for the nth derivative,

\frac{d^{n}}{dx^{n}} f(x).

Prove that your guess is right.