Chapter 11.5, Problem 15P

To determine

To find: the special kind of number that is formed by adding $1$ to the eight times triangular number.

Answer to Problem 15P

Perfect square ${\left(2n+1\right)}^2$ .

Explanation of Solution

Given:

Plutarch observed $1$ more than eight times a triangular number almost 2000 year ago.

Concept used:

Triangular number sequence is:

  $1,3,6,10,\mathrm{15......},\frac{n\left(n+1\right)}{2}$ .

The summation of triangular number is:

  $T_n={\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}.$

Algebraic identities:

  $a^2+2ab+b^2={\left(a+b\right)}^2$ .

Calculation:

The sums of the $nth$ triangular number is given by the summation formula:

  $T_n={\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}.$

Multiplying the summation formula by $8$ and add $1$ .

After simplifying, it shows that $8T_n+1$ produces a square number.

  $\begin{array}{l}8T_n+1=8\left(\frac{n\left(n+1\right)}{2}\right)+1.\\ \Rightarrow 4n\left(n+1\right)+1.\\ \Rightarrow 4n^2+4n+1.\\ \Rightarrow {\left(2n+1\right)}^2.\end{array}$

Hence, the solution is in perfect square ${\left(2n+1\right)}^2$ .