Recently I've been doing some work with trianglular numbers. Basically I wanted to show that for every nth triangular number $T_n$ $$T_n=\frac{n(n+1)}2$$ For me the simplicity if this equation is equal to the difficulty that requires to find it's demonstration (at least by me, I'm sure that the demonstration is really simple but I couldn't find it). I tried using induction but I failed miserably; but I know that the equation is true because I noticed it while drawing triangles and noticing that summing two triangular numbers they always form a square with base $n$. Basically I noticed that $$T_n+T_{n-1}=n^2 ;\ T_n = T_{n-1}+n \rightarrow T_n=\frac{n(n+1)}2$$ The problem can be reduced to only proving that $$T_n+T_{n-1}=n^2$$
Any idea how to prove the equation?
Use induction on $n$ and observe that the statement for $n=1$ is clear and suppose that we have $T(n)=\frac{n(n+1)}2$ for some natural number $n$, our aim is to prove that $T(n+1)=\frac{(n+1)(n+2)}2$ (induction step). For we have $$ T(n+1)=T(n)+n+1=\frac{n(n+1)}2+n+1=\frac{(n+1)(n+2)}2 $$ as we wanted.