Mathematical Induction

The mathematical induction is one of the numerical proof naturally it is used to find the given statement is accurate for all normal numbers. It can be a deducted by proving that the 1^st statement may be the countless succession of the statement is accurate and we have to prove that if any other statement in the countless succession is real. In this article we shall discuss about the examples involved in induction proof.

1. illustrate the true for n = 1

2. Suppose true for n = k

3. Explain true for n = k + 1

4. Conclusion: Statement is true for all n ≥ 1

Problem 1: Apply mathematical induction rule to prove 1 + 2 + 3 +… + n = n $\frac{(n+1)}{2}$ for all positive integers n.

Solution: Let p (n) be
1 + 2 + 3 +… + n = n $\frac{(n+1)}{2}$

Step 1: First we show p (1) is true
Left side = 1
Right side = 1 $\frac{(1+1)}{2}$ = 1
Both sides are equal hence p (1) is true

Step 2: Let us assume p (k) is true
1 + 2 + 3 +… + k = k $\frac{(k+1)}{2}$
Show that p (k+1) is true by adding a (k +1) on both the sides at the above statement
1 + 2 + 3 +… + k (k+1) = k $\frac{(k+1)}{2}$ + (k+1)
= (k+1) ($\frac{k}{2}$ + 1)

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

The above statement can be rewritten as
1 + 2 + 3 +… + k (k+1) = $\frac{(k+1)(k+2)}{2}$

This is the statement for p (k+1)

Problem 2: Apply mathematical induction rule to prove 1³ + 2³ + 3³+….+ n³ = n² $\frac{(n+1)^2}{4}$ for all positive integers n.

Solution: Let p(n) be

1³ + 2³ + 3³+….+ n³ = n² $\frac{(n+1)^2}{4}$

Step 1: First we show p (1) is true.

Left side = 1³ = 1

Right side = 1² $\frac{(1+1)^2}{4}$ = 1

Both sides are equal hence p (1) is true.

Step 2: Let us assume p (k) is true

1³ + 2³ + 3³+….+ k³ = k² $\frac{(k+1)^2}{4}$

Adding (k+1)³ on both the sides

1³ + 2³ + 3³+….+ k³ (k+1)³ =k² $\frac{(k+1)^2}{4}$ + (k+1)³

Factor (k+1)² on the right side

= (k+1)² [$\frac{k^2}{4}$ + (k+1)]

Now set a common denominator and make group

= (k+1)² $\frac{[k^2 + 4k + 4]}{4}$

= (k+1)² $\frac{[ (k+2)^2]}{4}$

The above statement can be rewritten as

1³ + 2³ + 3³+….+ k³ (k+1)³ = (k+1)² $\frac{[ (k+2)^2]}{4}$

This is the statement for p (k+1)