Chapter 2.6, Problem 24E

a.

To determine

To prove:The nth pyramid number $p_n$ is

  $p_n=1^2+2^2+\dots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

for every natural number n .

Explanation of Solution

Given information:

The first few pyramid number are $p_1=1,p_2=2,p_3=14,p_4=30,p_5=55$ .

Formula used:Use the induction hypothesis such that if a statement is true for $n=1$ then that statement is true for all natural number $n=k$ .

Proof:

Since, $p_1=1=1^2$

  $\begin{array}{l}{p}_2=5=1+4=1^2+2^2\\ p_3=14=1+4+9=1^2+2^2+3^2\\ \dots \dots \dots \dots \dots \\ \dots \dots \dots \dots \dots \end{array}$

Consider the result is true for $\left(n-1\right)$ that is

  $p_{n-1}=1^2+2^2+3^2+\dots +{\left(n-1\right)}^2$

Now assume the base is $n\times n$ squares. The layer above the base consists of $\left(n-1\right)\times \left(n-1\right)$ square.

Therefore, it is a pyramid with base $\left(n-1\right)\times \left(n-1\right)$ square above the given base $n\times n$ square.

  $\begin{array}{c}{p}_n=n^2+p_{n-1}\\ p_n=1^2+2^2+3^2+\dots +{\left(n-1\right)}^2+n^2\end{array}$

for every natural number n .

From the given statement,

  $p_n=1^2+2^2+\dots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\dots \left(1\right)$

Then,

  $\begin{array}{l}{p}_1=1=\frac{1\left(1+1\right)\left(2+1\right)}{6}\\ 1=1\end{array}$

This statement is true for $n=1$ .

Consider the statement (1) is true for some natural number $n=k$ .

Then,

  $p_k=1^2+2^2+\dots +k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\dots \left(2\right)$

Now prove that statement (2) is true $k=k+1$ .

take

  $1^2+2^2+3^2+\dots +k^2+{\left(k+1\right)}^2=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$

And

  $\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)\left(2\left(k+1\right)+1\right)}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$

Therefore, statement (2) is true for $k=k+1$ .

Hence, by induction hypothesis the statement is true for all-natural number n .

b.

To determine

To prove:The nth pyramid number $p_n$ is

  $p_n=\left(\begin{array}{c}n+2\\ 3\end{array}\right)+\left(\begin{array}{c}n+1\\ 3\end{array}\right)\text{for}n\ge 2$

Explanation of Solution

Given information:

The first few pyramid number are $p_1=1,p_2=2,p_3=14,p_4=30,p_5=55$ .

Proof:

Consider for $n\ge 2$ ,

  $\begin{array}{c}\left(\begin{array}{c}n+2\\ 3\end{array}\right)+\left(\begin{array}{c}n+1\\ 3\end{array}\right)=\frac{\left(n+2\right)!}{3!\left(n+2-3\right)!}+\frac{\left(n+1\right)!}{3!\left(n+1-3\right)!}\\ =\frac{\left(n+2\right)!}{3!\left(n-1\right)!}+\frac{\left(n+1\right)!}{3!\left(n-2\right)!}\\ =\frac{\left(n+2\right)\left(n+1\right)n\left(n-1\right)!}{3!\left(n-1\right)!}+\frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)!}{3!\left(n-2\right)!}\\ =\frac{\left(n+2\right)\left(n+1\right)n}{3!}+\frac{\left(n+1\right)n\left(n-1\right)}{3!}\\ =\frac{n\left(n+1\right)}{3\times 2}\left[n+2+n-1\right]\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Therefore,

  $\left(\begin{array}{c}n+2\\ 3\end{array}\right)+\left(\begin{array}{c}n+1\\ 3\end{array}\right)=p_n$

c.

To determine

To prove:The nth pyramid number $p_n$ is

  $p_n=\frac{1}{4}\left(\begin{array}{c}2n+2\\ 3\end{array}\right)$

for every natural number n .

Explanation of Solution

Given information:

The first few pyramid number are $p_1=1,p_2=2,p_3=14,p_4=30,p_5=55$ .

Proof:

Consider for every natural number n ,

  $\begin{array}{c}\frac{1}{4}\left(\begin{array}{c}2n+2\\ 3\end{array}\right)=\frac{1}{4}\frac{\left(2n+2\right)!}{3!\left(2n+2-3\right)!}\\ =\frac{1}{4}\frac{\left(2n+2\right)!}{3!\left(2n-1\right)!}\\ =\frac{1}{4}\frac{\left(2n+2\right)\left(2n+1\right)\left(2n\right)}{6}\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Therefore,

  $\frac{1}{4}\left(\begin{array}{c}2n+2\\ 3\end{array}\right)=p_n$

d.

To determine

To prove:The nth pyramid number $p_n$ is

  $p_n=n\left(\begin{array}{c}n+1\\ 2\end{array}\right)-\left(\begin{array}{c}n+1\\ 3\end{array}\right);\text{for}n\ge 2$

Explanation of Solution

Given information:

The first few pyramid number are $p_1=1,p_2=2,p_3=14,p_4=30,p_5=55$ .

Proof:

Consider $\text{for}n\ge 2$ ,

  $\begin{array}{c}n\left(\begin{array}{c}n+1\\ 2\end{array}\right)-\left(\begin{array}{c}n+1\\ 3\end{array}\right)=n\frac{\left(n+1\right)!}{2!\left(n+1-2\right)!}-\frac{\left(n+1\right)!}{3!\left(n+1-3\right)!}\\ =\frac{n\left(n+1\right)!}{2!\left(n-1\right)!}-\frac{\left(n+1\right)!}{3!\left(n-2\right)!}\\ =\frac{n\left(n+1\right)n}{2}-\frac{\left(n+1\right)n\left(n-1\right)}{6}\\ =\frac{n\left(n+1\right)}{6}\left[3n-n+1\right]\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Therefore,

  $n\left(\begin{array}{c}n+1\\ 2\end{array}\right)-\left(\begin{array}{c}n+1\\ 3\end{array}\right)=p_n$