Let $t(n)$ denote the n$^{th}$ triangular number.
Let $T(n)$ denote the sum of the first $n$ triangular numbers.
Is there a formula for $T(n)$?
$\endgroup$ 25 Answers
$\begingroup$Formula for T(n): $$T(n)=\frac{n(n+1)(n+2)}{6}$$ Proof: we can prove it in an inductive way.
Base case:$$T(1)=\frac{1*2*3}{6}=1$$ Let n=k. We have $T(n+1)=T(n)+t(n+1).$
Therefore $$T(k+1)=T(k)+t(k+1)=\frac{k(k+1)(k+2)}{6}+\frac{(k+1)(k+2)}{2}=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{6}=\frac{(k+1)(k+2)(k+3)}{6}$$
Q.E.D
$\endgroup$ 4 $\begingroup$$nth$ triangular number is the sum of $n$ consecutive natural numbers from starting which is simply $n(n+1)/2$. You want sum of first $n$ triangular numbers. Just take the sum $\Sigma_{i=1}^n \frac{i(i+1)}{2}$.
$\endgroup$ 3 $\begingroup$Since the $k$-th triangular number is $T(k)=\frac{k(k+1)}{2}$, so your sum is $$ \sum_{k=1}^n\frac{k(k+1)}{2}=\frac{1}{2}\biggl( \sum_{k=1}^n k^2+\sum_{k=1}^n k\biggr) $$ The second summation is $T(n)$, the first summation is $$ \frac{1}{3}n\left(n+\frac{1}{2}\right)(n+1) $$ (a nice way to memorize it), you find it in several places (the book “Concrete Mathematics” by Graham, Knuth and Patashnik features several derivations of the formula).
$\endgroup$ $\begingroup$$$\begin{align} t(r)&=\binom {r+1}2&&=\frac {r(r+1)}2\\ T(n)&=\sum_{r=1}^n t(r)\\ &=\sum_{r=1}^n \binom {r+1}2\\ &=\binom {n+2}3&&=\frac {n(n+1)(n+2)}6\end{align}$$
$\endgroup$ 4 $\begingroup$$$\sum_{k=1}^n\frac{k(k+1)}{2}=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)=$$ $$=\frac{n(n+1)}{12}\cdot(2n+1+3)=\frac{n(n+1)(n+2)}{6}.$$
I used that $\frac{n(n+1)}{2}$ is the $n$-th triangle number.
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