Question

Prove the following using mathematical induction:
1. (ab)n = an bn for every natural number n
2. 13 +23+33+…+n3 = (1+2+3+…+n)2
3. 1+3+5+7+…+(2n-1) = n2

Answer 1

Answer on Question #85286 - Programming & Computer Science - Algorithms

Question 85286:

Prove the following using mathematical induction:

1. $(ab)^n = a^n b^n$ for every natural number $n$

2. $1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = (1 + 2 + 3 + \dots + n)^{2}$

3. $1 + 3 + 5 + 7 + \dots + (2n - 1) = n^{2}$

Answer:

1. $(ab)^{1} - a^{1}b^{1} = 0$

$$(ab)^{n+1} - a^{n+1}b^{n+1} = ab(ab)^{n} - aa^{n}bb^{n} = ab((ab)^{n} - a^{n}b^{n}) = ab * 0 = 0$$

2. $1^{3} - 1^{2} = 0$

$$\begin{array}{l} 1^{3} + 2^{3} + 3^{3} + \dots + (n + 1)^{3} - (1 + 2 + 3 + \dots + (n + 1))^{2} = ((1^{3} + 2^{3} + 3^{3} + \dots + n^{3}) + \\ 1 + n^{3} + 3 * n^{2} + 3n) - ((1 + 2 + 3 + \dots + n)^{2} - (n + 1)^{2} - 2(n + 1)(1 + 2 + 3 + \dots + \\ n)) = 1 + n^{3} + 3 * n^{2} + 3n - (n + 1)^{2} - 2(n + 1)(1 + 2 + 3 + \dots + n) = 1 + n^{3} + 3 * n^{2} + \\ 3n - n^{2} - 2n - 1 - 2(n + 1)n(n + 1)/2 = n^{3} + 2 * n^{2} + n - (n + 1)^{2}n = 0 \end{array}$$

3. $(2 * 1 - 1) - 1^{2} = 0$

$$\begin{array}{l} 1 + 3 + 5 + 7 + \dots + (2(n + 1) - 1) - (n + 1)^{2} = 1 + 3 + 5 + 7 + \dots + (2n - 1) + \\ (2n + 2 - 1) - n^{2} - 2n - 1 = (2n + 2 - 1) - 2n - 1 = 0 \end{array}$$

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