The other tutor just gave you the answer. He didn't show how
to get it.

1,4,11,26,57,120,...}?

a1 = 1, a2 = 4, a3 = 11, a4 = 26, ...

For a recursion formula, let's try a linear combination 
of the previous term, an-1, the number of
the term, n, and a constant.  We'll let the coefficients be p, q,
and the constant be r:

an = p·an-1 + q·n + r

For n = 2

an = p·an-1 + q·n + r
a2 = p·a2-1 + q·2 + r
a2 = p·a1 + q·2 + r
4 = p·1 + 2q + r
4 = p + 2q + r

For n = 3

an = p·an-1 + q·n + r
a3 = p·a3-1 + q·3 + r
a3 = p·a2 + q·3 + r
11 = p·4 + 3q + r
11 = 4p + 3q + r

For n = 4

an = p·an-1 + q·n + r
a4 = p·a4-1 + q·4 + r
a4 = p·a3 + q·4 + r
26 = p·11 + 4q + r
26 = 11p + 4q + r



So we solve this system of equations:

 4 =   p + 2q + r
11 =  4p + 3q + r
26 = 11p + 4q + r

and get p=2, q=1, r=0

So the recursion becomes:

an = p·an-1 + q·n + r
an = 2·an-1 + 1·n + 0

or

an = 2an-1 + n 

Now we check to see if this recursion holds
for the remaining given terms:

2(11) + 4 =  22 + 4 =  26  yes
2(26) + 5 =  52 + 5 =  57  yes
2(57) + 6 = 114 + 6 = 120  yes

Indeed it does, so the next term is

      2(120) + 7 = 240 + 7 = 247

Edwin