Please help!!!

Good work guest!

I'll just show you how to do it formally.

Vinny wrote down all the single-digit base-$b$ numbers and added them in base $b$, getting $34_b$.What is $b$?

let the base be b then

1+2+3+.....(b-1)= 3b+4

This as an AP   

S= (b/2)(1+b) 

so

\(T_1=0,\qquad T_b=b-1\\ \frac{b}{2}(0+b-1)=3b+4\\b(b-1)=6b+8\\ b^2-b-6b-8=0\\ b^2-7b-8=0\\ (b-8)(b+1)\\ b=-1\;\;or\;\;8 \)

But b must be a possitive integer so the base is 8

Hmm, I give a take at this.

0+1+2+3+4+5+6+7 =28 base 10

28/8 =3 + 4 as a remainder

so, the answer is 34 base 8.