Day: September 29, 2020

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202009290227 Exercise 2.3.5

Prove that 10^{n+1}+10^n+1 is divisible by 3 for n\in\mathbb{N}.

Extracted from T. W. Judson. (2021). Abstract Algebra Theory and Applications.

Proof.

Let P(n) be the statement:

P(n):\qquad 3\,\Big| (10^{n+1}+10^n+1) for any n\in\mathbb{N}

Determine whether or not P(n) is true when n=1:

P(1): \qquad 3\,\Big| (10^{(1)+1}+10^{(1)}+1)

As 10^{(1)+1}+10^{(1)}+1 = 111 = 37\cdot 3 is divisible by 3, P(1) is true.

Suppose P(n) is true for some n\, (\geqslant 1) \in\mathbb{N}, try and prove the statement P(n+1):

P(n+1): \qquad 3\,\Big| (10^{(n+1)+1}+10^{(n+1)}+1)

\begin{aligned} &\quad 10^{(n+1)+1}+10^{(n+1)}+1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +10 - 9 \\ = &\quad 10\cdot (10^{n+1}+10^n+1) - 9 \\ \end{aligned}

As P(n) is true and by the fact that three divides nine, 10\cdot (10^{n+1}+10^n+1) - 9 is therefore divisible by 3. That is,

P(n)\textrm{\enspace is true\enspace}\Rightarrow P(n+1)\textrm{\enspace is true\enspace}

That P(n) is true for n=1, by the principle of mathematical induction, I have thus proven 10^{n+1}+10^n+1 is divisible by 3 for n\in\mathbb{N}.

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202009290114 Exercise 2.3.1

Prove that

1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}

for n\in\mathbb{N}.

Proof.

Let P(n) be the statement

1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}} for n\in\mathbb{N}.

When n=1, check for the validity of P(1):

\begin{aligned} \textrm{LHS} & = 1^2 = 1 \\ \textrm{RHS} & = \frac{(1)\big( (1)+1\big) \big( 2(1)+1\big) }{6} = 1 \\ \end{aligned}

\because \textrm{LHS}=\textrm{RHS}

\therefore P(n) is true for n=1.

Suppose P(n) holds true, let’s see if P(n+1) holds too:

P(n+1):\qquad 1^2+2^2+\cdots + n^2 + (n+1)^2 \stackrel{?}{=} \displaystyle{\frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6}}

Beginning with the left hand side,

\begin{aligned} \textrm{LHS} & = 1^2+2^2+\cdots + n^2 + (n+1)^2 \\ & = P(n) + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\ & = \frac{(n+1)\big[ (n)(2n+1) + 6(n+1) \big]}{6} \\ & = \frac{(n+1)(2n^2+n+6n+6)}{6} \\ & = \frac{(n+1)(2n^2+7n+6)}{6} \\ & = \frac{(n+1)(2n+3)(n+2)}{6} \\ \end{aligned}

Then turn to the right hand side,

\begin{aligned} \textrm{RHS} & = \frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6} \\ & = \frac{(n+1)(n+2)(2n+3)}{6} \\ \end{aligned}

\because \textrm{LHS} = \textrm{RHS}

\therefore P(n+1) holds when P(n) holds.

As is proven P(1) is true, by the principle of mathematical induction, P(n) is also true for n\geqslant 1, i.e.,

\forall\, n\in\mathbb{N},\qquad 1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}