This kind of multiple-choice question is really difficult to write. Generally speaking, it is good to use special values.

M is only 3 or 4. I'll take 3 n first. As long as it's greater than 4, I'll take a 5.

Then the formula becomes [(a1+a2+a3+a4+a5+a6+a7+A8)+(a1+a2)]/[(a1+a2+a3+a5)+(a/kloc)

(2a 1+2 a2+a3+a4+a5+a6+a7+A8)/(2a 1+2 a2+2 a3+a4+a5)= 2

2a 1+2 a2+a3+a4+a5+a6+a7+A8 = 4a 1+4a 2+4a 3+2 a4+2 a5

a6+a7+A8 = 2a 1+2 a2+3 a3+a4+a5

Bring the recipe in. The first two seem to be more troublesome. Take the last one first.

c:3n-2 16+ 19+22 = 2+8+2 1+ 10+ 13？ exclude

d:2n- 1 1 1+ 13+ 15 = 2+6+ 15+7+9？ reserve

The answer is D.

Why can't you think about the first two N powers, whether it's the eighth power of 3 or the eighth power of 2, the values are too big, and it's not enough to add up the single digits.

N times of A: 3-2 727+2185+6559 = 6+18+81+81+243? Hmm. How interesting

B: Degree of 2-163+127+255 n = 2+6+31+15+31? Hmm. How interesting

So choose d