Show that $u_{n+1}=\frac{(n+1)(n+2)}{2}$

Question

Given the sequence $u_1=1$ and $u_{n+1}=u_n+(n+1)$, show that $u_{n+1}=\frac{(n+1)(n+2)}{2}$.

This is what I have so far,

I know the sequence for this is 1, 3, 6, 10, 15, 21, 28 and

$u_n+(n+1)=\frac{(n+1)(n+2)}{2}$

To derive what $u_n$ should equal to, I noticed that because the second difference is the same in the sequence, I have to write a quadratic function ($ax^2+bx+c$. So I set up the system of the equation as follows: $a+b+c=1$; $4a+2b+c=3$; $9a+3b+c=6$

I found that $a=\frac{1}{2}, b=\frac{1}{2}, c=0$, and then I simplified and showed that $u_{n+1}=\frac{(n+1)(n+2)}{2}$

Is there another way or possibly a faster way to show that $u_{n+1}=\frac{(n+1)(n+2)}{2}$? Any help is appreciated.

this is trivial by induction – David C. Ullrich Jul 15 '21 at 15:48 — David C. Ullrich, Jul 15 '21 at 15:48
Telescoping like this: $u_{n+1}-u_n$ should also work. – Koro Jul 15 '21 at 15:50 — Koro, Jul 15 '21 at 15:50

score 1 · Answer 1 · answered Jul 15 '21 at 16:09

1

$u_{n+1} - u_n = n+1$

Thus we have $u_2-u_1= 1+1, u_3-u_2= 2+1,\ldots u_{n+1}-u_n= n+1$

Summing up we have $u_{n+1}-u_1= (1+2+3+...)+n\\ u_{n+1}= (1+n) + \dfrac{n(n+1)}{2}$

answered Jul 15 '21 at 16:09

Subham Karmakar

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Show that $u_{n+1}=\frac{(n+1)(n+2)}{2}$

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