Question

Question one

Prove by induction that the following statements are true for all integers n
a) 12x2+22x3+…..+n2(n+1)= n(n+1)(n+2)(3n+1)/12
b)4007n-1 is divisible by 2003

Answer 1

Answer on Question #75987 – Math – Discrete Mathematics

Question

Prove by induction that the following statements are true for all integers $n$

a) $1^2 \times 2 + 2^2 \times 3 + \dots + n^2(n + 1) = n(n + 1)(n + 2)(3n + 1)/12$

b) $4007^n - 1$ is divisible by 2003

Solution

a) $1^2 \times 2 + 2^2 \times 3 + \dots + n^2(n + 1) = n(n + 1)(n + 2)(3n + 1)/12$

Let's prove this equality by induction on $n$:

1. $n = 1$: $2 = 1^2 \times 2 = \frac{1 \times 2 \times 3 \times 4}{12} = \frac{24}{12} = 2$.

2. Assume that this equality holds for $n = k$:

$$1^2 \times 2 + 2^2 \times 3 + \dots + k^2(k + 1) = k(k + 1)(k + 2)(3k + 1)/12$$

3. Let's prove equality for $n = k + 1$.

$$\begin{aligned} 1^2 \times 2 + \dots + (k + 1)^2((k + 1) + 1) &= 1^2 \times 2 + \dots + (k^2 + 2k + 1)(k + 2) = \\ &= 1^2 \times 2 + \dots + k^3 + 4k^2 + 5k + 2 = \frac{k(k + 1)(k + 2)(3k + 1)}{12} + k^3 + 4k^2 + \\ &+ 5k + 2 = \frac{k(k + 1)(k + 2)(3k + 1) + 12k^3 + 48k^2 + 60k + 24}{12} = \\ &= \frac{((k + 1) + 1)((k + 1) + 2)((k + 1) + 3)(3(k + 1) + 1)}{12} \Rightarrow \end{aligned}$$

$\Rightarrow$ the equality is proved.

The statement is true for all natural $n$ by the principle of mathematical induction.

b) $4007^n - 1 : 2003$

Let's prove this equality by induction on $n$:

1. $n = 1$: $4007^1 - 1 = 4006 : 2003$.

2. Assume that this equality holds for $n = k$: $4007^k - 1 : 2003$.

3. Let's prove equality for $n = k + 1$.

$$4007^{k+1} - 1 = 4007 \times 4007^k - 1 = 4006 \times 4007^k + 4007^k - 1.$$

1) $4006 \times 4007^k + 4007^k : 2003$;

2) $4007^k - 1 : 2003 \Rightarrow 4006 \times 4007^k + 4007^k - 1 : 2003$.

$\Rightarrow$ the formula is proved.

The statement is true for all natural $n$ by the principle of mathematical induction.

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