Solve the recurrence relation:

Question

anonymous · Answer

$$ a_{n} = a_{n-1} + n, a_{0} = 1$$

loser66 · Answer

solve what? just plug n =1, 2, 3.... in.

loser66 · Answer

oh, solve for the formula, right? wow... it was 3 years ago when I took this course. hehehe...

anonymous · Answer

yeah

anonymous · Answer

glad you still got it

anonymous · Answer

i know they come out to be 2, 4, 7 but I don't see no pattern there with exponents or anything..... ugh

anonymous · Answer

no your reply!

loser66 · Answer

$$a_1 = a_0+1\a_2= a_1+2= a_0+1+2=a_0+3\a_3= a_2+3= a_0+3+3=a_0+6\.......\a_n=???$$

loser66 · Answer

All you have to do is interpret the number at the end of each row. You see, at row 2, that is $a_\color{red}{2}$, then the last number is 3
row 3, $a_\color{blue}{3}$ then the last number is 6
do some more, you can construct the link between them. 
Note: the first term is $a_0$ always, no need to look at it.

anonymous · Answer

then $a_{4} = a_{0} + 10 $ and i don't see a link

loser66 · Answer

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