Problem 16

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer $n.$ $$2+7+12+\dots+(5 n-3)=\frac{n(5 n-1)}{2}$$

Step-by-Step Solution

Verified

Answer

The given statement is proven to be true by mathematical induction. We successfully showed the base case that the proposition holds for $n=1$. By assuming it holds for $k$ and showing it also holds for $k+1$, we have completed the induction step. Therefore, the statement $2+7+12+\dots+(5 n-3)=\frac{n(5 n-1)}{2}$ is shown to be true for all positive integers $n$.

1Step 1: Base Case

Check if the given formula holds true for $n=1$. The left-hand side of the equation equals to $2$, and the right-hand side equals to $\frac{1*(5*1-1)}{2} = 2$. The equation holds for n = 1, so the base case is verified.

2Step 2: Inductive Step

Assume the formula holds for some positive integer $k$, i.e. $2+7+12+\dots+(5 k-3) = \frac{k(5 k-1)}{2}$. Now, to prove it for $k+1$, examine the left-hand side and add the next term. The new sum becomes $2+7+12+\dots+(5 k-3) + (5(k+1) - 3)$ which equals to $\frac{k(5 k-1)}{2} + (5(k+1) - 3)$.

3Step 3: Simplify

Simplify the equation from Step 2. It should look like this: $\frac{k(5k-1)}{2} + 5k + 2 = \frac{2k^2+3k+2}{2}$. The right side of the equation we want to prove equals to $\frac{(k+1)(5(k+1)-1)}{2} = \frac{2k^2+3k+2}{2}$. Both sides are equal, thus proving the induction step.

Key Concepts

Base caseInductive stepProof by inductionAlgebraic proof

Base case

In mathematical induction, the base case is the starting point of the proof. It demonstrates that the statement to be proven works for the initial value of the variable, usually when $ n = 1 $. This step is crucial because it establishes the validity of the statement at the beginning of the sequence. If the base case does not hold, the entire proof by induction falls apart.

For our exercise, the base case requires us to check if the formula $ 2 + 7 + 12 + \dots + (5n-3) = \frac{n(5n-1)}{2} $ is true when $ n = 1 $.
On the left-hand side, substituting $ n = 1 $ gives us 2.
On the right-hand side, $ \frac{1(5 \times 1 - 1)}{2} = 2 $.

Both sides equal 2, confirming that the statement is true when $ n = 1 $. Therefore, the base case is successfully verified, serving as a solid foundation for the inductive proof.

Inductive step

The inductive step is the heart of a proof by induction. It involves two parts: the assumption and the induction. This step proves that if a statement holds for an arbitrary positive integer $ k $, it must also hold for the next integer, $ k+1 $. By doing so, the domino effect of sorts is set in motion.

First, assume the statement is true for a particular $ k $. For our example, assume $ 2 + 7 + 12 + \dots + (5k-3) = \frac{k(5k-1)}{2} $.
Then, prove it for $ k+1 $ by adding the next term to both sides: $ 2 + 7 + 12 + \dots + (5k-3) + (5(k+1) - 3) $.
This becomes $ \frac{k(5k-1)}{2} + (5(k+1) - 3) = \frac{2k^2+3k+2}{2} $.

The step involves algebraic manipulation until both sides of the equation match. Once they do, it establishes that the statement is true for $ k+1 $, thereby completing the inductive step.

Proof by induction

Proof by induction is a powerful method used in mathematics to show that a statement or formula is true for all natural numbers. It is akin to proving a line of dominoes will fall over, starting with the first piece.

It consists of two main parts: verifying the base case and demonstrating the inductive step.
The base case establishes the truth of the statement for the initial value, while the inductive step extends this truth to all subsequent values.
In the exercise provided, proof by induction is employed effectively to show the truth of the given summation formula for all positive integers $ n $.

By completing both the base case and inductive step successfully, proof by induction allows us to conclude with certainty that the statement holds true for any value of $ n $, not just a few specific cases.

Algebraic proof

An algebraic proof within the context of induction often involves manipulating algebraic expressions to demonstrate the truth of a mathematical statement. This requires a solid grasp of algebraic techniques to simplify and work with expressions.

In the inductive step, algebraic proof is used to transition from the assumption for $ k $ to proving the statement for $ k+1 $.
The process includes adding the new term to both sides of the equation and simplifying the expressions to show that they are equal.
For the exercise, after assuming $ 2 + 7 + 12 + \dots + (5k-3) = \frac{k(5k-1)}{2} $, the expression $ 5(k+1) - 3 $ is added, and careful algebraic simplification is necessary to achieve the result: $ \frac{2k^2+3k+2}{2} $.

Algebraic proof taps into the power of algebra to validate mathematical statements, providing a logical and rigorous framework for induction.

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