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 $T(n)$?
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10I don't see an obvious reason to create such a tag. Could you elaborate why you think triangular numbers need a separate tag? – TZakrevskiy Sep 19 '17 at 10:05
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4For reference: you want the tetrahedral numbers. – J. M. ain't a mathematician Sep 19 '17 at 13:56
5 Answers
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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
abc...
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Strictly speaking, this proof isn't valid because it uses the formula it's trying to prove to establish the base case. Better to simply compute the "sum" of the first triangular number explicitly. – David Z Sep 19 '17 at 15:22
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@DavidZ: I take your point, but I read that line as stating that $T(1)$ (ie the first triangular number) is equal to ${1\cdot 2 \cdot 3}\over 6$ (ie formula we're trying to prove) - so no assumption is necessarily being made. – psmears Sep 19 '17 at 16:43
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@psmears Oh, OK, I see what you mean. Well, in that case I would scale back my criticism to merely say that it's not explained well. – David Z Sep 19 '17 at 16:47
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$nth$ 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 $\Sigma_{i=1}^n \frac{i(i+1)}{2}$.
Vidyanshu Mishra
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2Actually you claimed something and then proved it. It is not always an easy task to make claims. – Vidyanshu Mishra Sep 19 '17 at 09:49
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$$\begin{align} t(r)&=\binom {r+1}2&&=\frac {r(r+1)}2\\ T(n)&=\sum_{r=1}^n t(r)\\ &=\sum_{r=1}^n \binom {r+1}2\\ &=\binom {n+2}3&&=\frac {n(n+1)(n+2)}6\end{align}$$
Hypergeometricx
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`pure' math(s) includes words, sometimes, and what's more this answer contradicts several others given here. – Sort of Damocles Sep 16 '18 at 16:39
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@dbx - No contradiction - the end result is the same i.e. $\binom {n+2}3$. The $3$ is the expanded form is a typo which has been corrected. – Hypergeometricx Sep 16 '18 at 17:34
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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).
egreg
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$$\sum_{k=1}^n\frac{k(k+1)}{2}=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)=$$ $$=\frac{n(n+1)}{12}\cdot(2n+1+3)=\frac{n(n+1)(n+2)}{6}.$$
I used that $\frac{n(n+1)}{2}$ is the $n$-th triangle number.
Michael Rozenberg
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