Recurrent equation mismatch in operator

Question

I am solving reccurent equation

$a_{n+1} = a_{n} + n + 1$

where a1 = 2.

which we can rewrite as

$a_{n} = a_{n-1} + n -1 + 1$

we can "expand" it such as

$ =a_{n-2} + n - 2 + 1 + n - 1 + 1$ $=a_{n-3} + n - 3 + 1 + n - 2 + 1 + n - 1 + 1$

we see that this is basicly sum of ( n - 1 ) to ( n - k ) , where k = n - 1.

Also we see , that 1 is added n - 1 times.

The formula according "Rules" i wrote should look like

$(\sum_{i=1}^{n-1}i ) + n - 1 $

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which we can rewrite as

$\frac{n^{2}+n-2}{2}$

but the correct answer according to wolfram should be

$\frac{n^{2}+n+2}{2}$

Which indicates that my though process was wrong. Where did i make mistake which lead to wrong result? Thanks

Answer 1

You correctly established that $a_i-a_{i-1}=i$. This relation holds for $i\geq 2$.

On the other hand you did not consider that $a_1=2$. In fact $$a_n=a_1+\sum_{i=2}^n (a_{i}-a_{i-1})=2+\sum_{i=2}^n i =2+\frac{n(n+1)}{2}-1=\frac{n^2+n+2}{2}.$$

Answer 2

write:

$$\sum_{i=1}^{n}(a_{i+1}-a_i)=\sum_{i=1}^{n}(i+1) \Rightarrow a_{n+1}-a_1=\frac{(n+3)(n)}{2} \Rightarrow a_{n+1}=2+\frac{(n+3)(n)}{2} \Rightarrow$$ $$a_n=2+\frac{(n+2)(n-1)}{2}=\frac{n^2+n+2}{2}$$

Answer 3

Basically, your idea is correct but there is a wrong calculation in that you did not add $a_1 =2$. $$a_{n+1} = a_n +n +1 $$ $$\Rightarrow a_n = a_{n-1} + (n-1) + 1 $$ $$\Rightarrow a_n = a_{n-2} + (n-2) + (n-1) + 1 + 1$$and so on. Hence, we can write $$ a_n = a_{n-k} +kn - \frac{k(k+1)}{2} + k$$ $$\Rightarrow a_n = a_{n-k} + kn-\frac{k(k-1)}{2}$$ When $k = n-1$, we have $$a_n = a_1 + n(n-1) -\frac{(n-1)(n-2)}{2}$$ $$\Rightarrow 2 +(n-1)[n-\frac{n-2}{2}]$$ $$\Rightarrow 2 + (n-1)[\frac{n+2}{2}]$$ $$ 2 + \frac{(n-1)(n+2)}{2} = \frac{n^2+n+2}{2}$$.