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I need to get the closed form of $1\cdot 2+2\cdot 3+3\cdot 4+\dots+n\cdot (n+1)$ I do know the ans is $n(n+1)(n+2)/3$ but i have no idea of the steps.

Here are some of my scratch works:

1 II(2)

2 IIIIII(6) difference of 4

3 IIIIIIIIIIII(12) difference of 6

............. (the pattern of difference observed is $2k$ for $k$ is 1 to $n$)

n IIIIIIII...........IIIIIII($n(n+1)$)

so i have Sum down vertical :

$$= 2n + 4(n-1) + 6(n-2) + ... +2n(n-(n-1))$$

after some simplify and grouping ,i got this:

$$2{ n\cdot n\cdot (n+1)/2 - (2+6+12+20+...+2n)}$$

I can't see i can get to the final result.

Great thanks for helping me out.

HaHaman
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4 Answers4

3

We have that $$1\cdot 2+2\cdot3+3\cdot 4+\dots+n\cdot(n+1)=2\sum_{k=2}^{n+1}\binom{k}{2}=2\binom{n+2}{3}=\frac{(n+2)(n+1)n}{3}.$$ where we used the Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$

Community
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Robert Z
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2

By indeterminate coefficients:

$S_n$ will be of the form $an^3+bn^2+cn+d$. Then solve for $S_0=0,S_1=2,S_2=8,S_3=22.$

It's even easier to start from

$$S_n-S_{n-1}=n(n+1)=a(3n^2-3n+1)+b(2n-1)+c$$ and identify the powers, giving

$$3a=1,-3a+2b=1,a+b-c=0.$$

2

Take the finite geometric series $1 + x + x^2 + \ldots +x^{n+1} = \frac{x^{n+2}-1}{x-1}$. Now differentiate both sides two times and take the limit $x \to 1$.

SlowLarry
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There's a few ways. Induction is probably the easiest. I think there's a way which involves assembling little cubes into a cuboid $n \times (n+1) \times (n+2)$, but that's basically just induction again and my drawing skills aren't good enough. (There might be an even more magic way which isn't fundamentally some kind of induction, but I don't see it.)

Patrick Stevens
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