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?

  • 1
    $\begingroup$ This is pretty straight-forward using induction. Maybe you should post your failed induction attempt (into your question) & we can show you where you went wrong. $\endgroup$ – PM 2Ring May 1 '15 at 10:42
  • $\begingroup$ Why somebody put the -1? $\endgroup$ – Peterix May 1 '15 at 11:00

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.

  • $\begingroup$ I don't understand why $T(n+1)=T(n)+n+1$ that was the main reason why why I had problem with induction $\endgroup$ – Peterix May 1 '15 at 10:54
  • $\begingroup$ This is the definition of triangular numbers. What's the definition in your sense? $\endgroup$ – k1.M May 1 '15 at 10:56
  • $\begingroup$ I just realised I'm stupid; thanks for the help man I thought that $T(n+1)=T(n)+n$ $\endgroup$ – Peterix May 1 '15 at 10:59
  • $\begingroup$ You're welcome. $\endgroup$ – k1.M May 1 '15 at 11:00

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