We want to show algebraically that any square number can be expressed as the sum of two consecutive triangular numbers.

Triangular Numbers:

The formula for the $n$-th triangular number, $T_n$, is:

$T_n = \frac{n(n+1)}{2}$

Statement to Prove:

We need to show that for any square number, $n^2$, it can be written as the sum of two consecutive triangular numbers. Specifically, we want to prove:

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

where $T_n$ is the $n$-th triangular number and $T_{n-1}$ is the $(n-1)$-th triangular number.

Step-by-Step Proof:

1. Write the formula for $T_n$ and $T_{n-1}$: Using the formula for the triangular numbers: $T_n = \frac{n(n+1)}{2}$ $T_{n-1} = \frac{(n-1)n}{2}$

2. Add the two triangular numbers: We now add $T_n$ and $T_{n-1}$: $T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{(n-1)n}{2}$

3. Simplify the expression: Combine the terms under a common denominator: $T_n + T_{n-1} = \frac{n(n+1) + (n-1)n}{2}$ Distribute $n$ inside both terms: $T_n + T_{n-1} = \frac{n^2 + n + n^2 - n}{2}$ Simplify further: $T_n + T_{n-1} = \frac{2n^2}{2}$

4. Final result: The right-hand side simplifies to: $T_n + T_{n-1} = n^2$

Thus, we have shown that any square number $n^2$ is indeed the sum of two consecutive triangular numbers.

Conclusion:

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

This completes the algebraic proof.

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Do you want further details or have any questions?

Here are 5 related questions that extend this concept:

1. How does the formula for triangular numbers relate to other polygonal numbers?
2. Can you prove this result geometrically?
3. What happens if you sum non-consecutive triangular numbers?
4. How can you use triangular numbers to find the sum of the first $n$ natural numbers?
5. How can this concept be generalized to other number patterns?

Tip: Whenever you are dealing with sums of numbers following a particular pattern, using their explicit formulas is a good way to start your proof.