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enter preformatted text here find the recurrence relation.

a(n)=a(n−1)+n with a(0)=0

Do I have to make a replace? Can someone help with initial steps?

Thanks.

Do like this....

https://scontent.fbkk12-1.fna.fbcdn.net/v/t1.15752-9/71552381_440510746590967_176859958611017728_n.jpg?_nc_cat=106&_nc_eui2=AeFjuLrF11yTRqm4rwRxkidD73Awi27JMpFJtOzJGrYFzcUKD45uCvjYSr9fw9LrD7DG3I7mPqUkFbMBzTa_Ow3RhCBwhvOamhkaedjXjgST7Q&_nc_oc=AQmj3z-oF-cM7g91fFyfcNd1H6SDpR9fo_G2aAXeINxQz6QcasHIekyarWmERqHHkQw&_nc_ht=scontent.fbkk12-1.fna&oh=49cb4d09bd7979bd9d394e561310dd89&oe=5E1E5D2C

enter image description here

Draasmae Awakap
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    This should be a very familiar sequence to you. $a(n)=n+a(n-1)=n+(n-1)+a(n-2)=\dots = n+(n-1)+(n-2)+\dots+3+2+1$. The search term here is "Triangle Numbers" – JMoravitz Oct 06 '19 at 17:57
  • Thank you very much but I want a solution like this please https://scontent.fbkk12-1.fna.fbcdn.net/v/t1.15752-9/71552381_440510746590967_176859958611017728_n.jpg?_nc_cat=106&_nc_eui2=AeFjuLrF11yTRqm4rwRxkidD73Awi27JMpFJtOzJGrYFzcUKD45uCvjYSr9fw9LrD7DG3I7mPqUkFbMBzTa_Ow3RhCBwhvOamhkaedjXjgST7Q&_nc_oc=AQmj3z-oF-cM7g91fFyfcNd1H6SDpR9fo_G2aAXeINxQz6QcasHIekyarWmERqHHkQw&_nc_ht=scontent.fbkk12-1.fna&oh=49cb4d09bd7979bd9d394e561310dd89&oe=5E1E5D2C – Draasmae Awakap Oct 06 '19 at 18:15
  • That IS a solution like that. But instead of multiplying be $2$ and getting $2^nC_0$ you multiply by $1$ and get $a_0$ and instead of adding $1$ one time you add $1 + 2+3 + ......+n$. Advice: Don't try to pound solutions into prefabricated examples without attempting to understand them. Attempt to see and understand how they work. – fleablood Oct 06 '19 at 18:28

1 Answers1

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

$a_{n-2}+ (n-1) + n=$

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

.....

$ a_2 + 3 + 4...... + (n-2) + (n-1) + n=$

$a_1 + 2 + 3 + 4...... + (n-2) + (n-1) + n=$

$a_0 + 1 + 2+ 3 + 4...... + (n-2) + (n-1) + n=$

$0+1 + 2+ 3 + 4...... + (n-2) + (n-1) + n=$

$\frac {n(n+1)}2$.

fleablood
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