Chapter 12.9, Problem 10CFU

a.

To determine

Find the next five triangular numbers.

Answer to Problem 10CFU

  $T_4=10,T_5=15,T_6=21,T_7=28,T_8=36$

Explanation of Solution

Given information:

Triangular numbers can be represented by a triangular array of dots, with $n$ dots on each side. The first three triangular numbers are $1,3,6$

  

Calculation:

Three triangular numbers are $1,3,6$

Next five numbers are,

  $\begin{array}{l}{T}_4=1+2+3+4\\ =10\\ T_5=1+2+3+4+5\\ =15\\ T_6=1+2+3+4+5+6\\ =21\\ T_7=1+2+3+4+5+6+7\\ =28\\ T_8=1+2+3+4+5+6+7+8\\ =36\end{array}$

b.

To determine

Write the general formula for $n^{th}$ sequence.

Answer to Problem 10CFU

  $\frac{n\left(n+1\right)}{2}$

Explanation of Solution

Given information:

Triangular numbers can be represented by a triangular array of dots, with $n$ dots on each side. The first three triangular numbers are $1,3,6$

  

Calculation:

Three triangular numbers are $1,3,6$

The $n^{th}$ term of triangular number is,

  $T_n=1+2+3+\mathrm{....}+n$

The sun of $n^{th}$ term of series contain first term $1$ and common difference $1$ .

  $\begin{array}{l}{T}_n=\frac{n}{2}\left(2\cdot a+\left(n-1\right)d\right)\\ T_n=\frac{n}{2}\left(2\cdot 1+\left(n-1\right)1\right)\\ T_n=\frac{n}{2}\left(2+n-1\right)\\ T_n=\frac{n\left(n+1\right)}{2}\end{array}$

Hence, the general formula for $n^{th}$ sequence is $\frac{n\left(n+1\right)}{2}$

c.

To determine

Prove that the sum of $n$ triangular numbers by $\frac{n\left(n+1\right)\left(n+2\right)}{6}$ .

Answer to Problem 10CFU

  $\frac{n\left(n+1\right)\left(n+2\right)}{6}$ Proved.

Explanation of Solution

Given information:

Triangular numbers can be represented by a triangular array of dots, with $n$ dots on each side. The first three triangular numbers are $1,3,6$

  

Calculation:

By using the principle of mathematical induction,

  $\begin{array}{l}{S}_n\equiv 1+3+6+\mathrm{..}+\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(n+2\right)}{6}\\ verify{S}_{n}isvalidforn=1\\ \frac{1\left(1+1\right)\left(1+2\right)}{6}=1\end{array}$

  $\begin{array}{l}assume{S}_{n}isvalidforn=k\\ 1+3+6+\mathrm{..}+\frac{k\left(k+1\right)}{2}=\frac{k\left(k+1\right)\left(k+2\right)}{6}\end{array}$

If the formula is valid for $n=k$ it also valid for $n=k+1$ since $S_n$ is valid for $n=1,n=2,n=3soon$

  $\begin{array}{l}1+3+6+\mathrm{..}+\frac{k\left(k+1\right)}{2}+\frac{\left(k+1\right)\left(k+2\right)}{2}=\frac{k\left(k+1\right)\left(k+2\right)}{6}+\frac{\left(k+1\right)\left(k+2\right)}{2}\\ =\frac{\left(k+1\right)\left(k+2\right)}{6}\left(k+3\right)\\ =\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\\ =\frac{\left(k+1\right)\left(\overline{k+1}+1\right)\left(\overline{k+1}+2\right)}{6}\end{array}$

Hence, $\frac{n\left(n+1\right)\left(n+2\right)}{6}$ proved.