各项均为正数的数列{an}，满足a1=1，a(n+1)方-an方=2 求通项公式，求数列{an方⼀2的n次方}的前n项和Sn

an方 - a(n-1)方 = 2
...
a2方 - a1方 =2
n-1个等式通加, 得
an方 - a1方 =2 *(n-1)
an方 = 2 *(n-1) + a1方 = 2n -1
因为各项均为正数 所以an = √(2n-1)

an方/2的n次方
= (2n-1)/2^n

设 bn = 2n-1, cn = 1/2^n
an方/2的n次方 = bn * cn
Sn = b1 *c1 + b2* c2 + ... + bn *cn
1/2 * Sn = b1 * c2 + b2 * c3 + ... + bn * c(n+1)
Sn - 1/2 *Sn = b1 *c1 + 2 *c2 + 2 *c3 + ... + 2 * cn - bn * c(n+1)
= b1 *c1 + 2 (c2 +c3 + ... +cn) + bn *c(n+1)
= 1*1/2 + 2* (1/2 - 1/2^(n+1))/(1- 1/2) + (2n-1) * 1/2^(n+1)
= 1/2 + 4*(1/2 - 1/2^(n+1))+ (2n-1) * 1/2^(n+1)
= 3/2 - 5 * 1/2^(n+1) + n *1/2^n

Sn = 3 - 5 *1/2^n + n * 1/2^(n-1)

a(n+1)方 “方” 的意思是什么意思