Multiply by the common ratio (or the reciprocal of the common ratio). This method can be used when a series is obtained by multiplying arithmetic progression and geometric progression. Your question, the front n is arithmetic progression, and the back (1/2) (n- 1) is geometric progression, so it is obtained by multiplying arithmetic progression and geometric progression.

Let an be arithmetic progression, bn be geometric progression, and cn = An BN (your topic is this type), then the sum of the first n items in cn.

The solution of Sn is usually multiplied by the common ratio Q of bn (or the reciprocal of common ratio 65438 +0/q), and then it can be solved by dislocation subtraction.

sn =(a 1 b 1)+(a2 B2)+……+(an bn)

Then q sn = (a1b1q)+(a2b2q)+...+(anbnq), and substitute b {n+ 1} = bn q, we get.

q Sn =(a 1 B2)+(a2 B3)+……+(a { n- 1 } bn)+(a b { n+ 1 })

Subtract two expressions to get.

( 1-q)Sn =(a 1 b 1)+(a2-a 1)B2+(a3-a2)B3+……+(an-a { n- 1 })bn-(an b { n+ 1 })，

Note that a2-a1= a3-a2 = ... = an-a {n-1} = d, so

(1-q) sn = (a1b1)+d (B2+B3+...+bn)-(anb {n+1}), and the middle bracket is the sum of geometric series, so sn can be obtained.

I won't give you the answer directly here. I'll give you a way. When you learn it, this kind of problem will be no problem.