A formula for the sum of the triangular numbers?

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


$nth$ 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 $\Sigma_{i=1}^n \frac{i(i+1)}{2}$.


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).

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