Question

a_n=a_n-1+n, a_0=2

Answer 1

using substitution method

Answer 2

a_n=a_n-1 + n

putting n=1

a_1 = a_1-1 + 1

a_1=a_0+1

a_1=2+1=3

like wise i hope it will be the series...

2+3+4+.... +a_n-1 + n = 1+2+3+4.... +a_n-1 +n -1 = n(n-1)/2 -1 = n^2 -n -2/2 = n^2 +n -2n-2/2 = n(n+1)-2(n+1)/2 = (n-2)(n+1)/2 thats the answer i hope/...

Answer 3

or one can write\[a_n-a_{n-1}=n\]\[a_{n-1}-a_{n-2}=n-1\]...\[a_1-a_{0}=1\]add all of them u have\[a_n-a_0=1+2+3+...+n=\frac{n(n+1)}{2}\]\[a_n=a_0+\frac{n(n+1)}{2}\]i hope its clear :)

Answer 4

\[(2) a _{n}-9a_{n}-1+26a_{n-2} =0 for n \ge3 \]