Chapter 9.4, Problem 51E

Solution

To Prove: The sum of two consecutive triangular numbers is a square number that is, $T_{n-1}+T_n=n^2$ for all positive integers $n\ge 3$ .

Given information:

The given statement says to prove that $T_{n-1}+T_n=n^2$ .

Explanation:

Let’s take the triangular number as: $1,3,6,10,15,\cdots$ . These are the numbers that count the object as triangular arrays.

The first five terms are: $T_1=1,T_2=3,T_3=6,T_4=10,T_5=15$

The terms can be written as:

  $\begin{array}{l}{T}_1=\frac{1}{2}\left(1^2+1\right)\\ T_2=\frac{1}{2}\left(2^2+2\right)\\ T_3=\frac{1}{2}\left(3^2+3\right)\\ T_4=\frac{1}{2}\left(4^2+4\right)\end{array}$

Then the nth term will become: $T_n=\frac{1}{2}\left(n^2+n\right)$

Similarly, the $\left(n-1\right)th$ term will become: $T_{n-1}=\frac{1}{2}\left({\left(n-1\right)}^2+\left(n-1\right)\right)$

Now the sum of nth and $\left(n-1\right)th$ term is:

  $\begin{array}{c}{T}_n+T_{n-1}=\frac{1}{2}\left(n^2+n\right)+\frac{1}{2}\left[{\left(n-1\right)}^2+\left(n-1\right)\right]\\ =\frac{1}{2}\left(n^2+n\right)+\frac{1}{2}\left[n^2+1-2n+n-1\right]\\ =\frac{1}{2}{n}^2+\frac{1}{2}n+\frac{1}{2}{n}^2-\frac{1}{2}n\\ =n^2\end{array}$

Therefore, $T_n+T_{n-1}=n^2$ .

Hence it is proved that the sum of two consecutive triangular numbers $T_n+T_{n-1}$ is $n^2$ .