General term formula for solving sequence 2, - 4, - 6, - 8

General term formula for solving sequence 2, - 4, - 6, - 8

-2n*(-1)^[2^(n-1)]
Where (- 1) ^ [2 ^ (n-1)] means the 2 ^ (n-1) power of (- 1)
Please substitute n = 1, 2, 3, 4 to verify. If correct, please adopt,

1.2.4.7.11.16

An-An-1=n-1

A】 Fill in the blanks according to the rule: 1, 1, 2, 3, 4, 5, 8, 7, 16, 9, 32, 11, 64, 13, 128, () -; b] write the general term formula, that is, the rule

A)15, 256
B) [(-1)^(n+1)+1]/2 * 2^[(n-1)/2] + [(-1)^n+1]/2 * (n-1)

Excuse me: find the general formula of sequence 4, 8, 13, 19, 26, 34
Find an =?

a1=4,a2=8、a3=13、a4=19、a5=26、a6=34
a2=a1+4
a3=a2+5
a4=a3+6
a5=a4+7
a6=a5+8
.
an=a(n-1)+n+2
Add all the formulas
an=a1+(4+5+...+(n+2))=4+(n-1)(n+6)/2
=(n^2+5n+2)/2