Step by step solution to "Use mathematical induction to prove that n(n2+5) is..."

High School
Algebra
Algebra I
Algebra foundations

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Answered question

2020-11-09

Use mathematical induction to prove that $n (n^{2} + 5)$ is divisible by 6 for $n \in N$

Answer & Explanation

d2saint0

Skilled2020-11-10Added 89 answers

Let P(n) denotes that $n (n^{n + 5})$ is divisible by 6, for each natural number n.
$P (1) : 1 (1^{2 + 5}) = 6$ , which is divisible by 6. Therefore the result is true for n=1.
Now, assume that P(n) is true for n=k. That is $P (k) : k (k^{2 + 5}) is \div i is \leq by 6 . k (k^{2} + 5) = 6 m$ for some m∈N. We have to prove that P(k+1) is also true.
$P (k + 1) : (k + 1) [{(k + 1)}^{2 + 5}]$

$= (k + 1) [k^{2} + 2 k + 1 + 5]$

$= (k + 1) [k^{2} + 2 k + 6]$

$= k^{3} + 2 k^{2} + 6 k + k^{2} \cdot 2 k + 6$

$= k^{3} + 3 k^{2} + 8 k + 6$

$= k^{3} + 5 k + 3 k^{2} + 3 k + 6$

$= k (k^{2} + 5) + 3 (k^{2} + k + 2)$

$= (6 m) + 3 (k^{2} + k + 2)$
Since $k^{2} + k + 2$ is divisible by 22 for every even and odd k. Therefore $3 (k^{2} + k + 2)$ is divisible by 66 and hence $(6 m) + 3 (k^{2} + k + 2)$ is divisible by 6. Therefore, $P (k + 1)$ is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.

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