Question

高二数学 数列求和！求过程两道题！

Answer 1

解：
大哥，你的字是用脚蹬的嘛？也太难认了吧！
因为看不清楚，猜着解吧！
1.
∵an=n · 2^(2n-1)
Sn = a1+a2+....+an
∴
Sn = 1·2 + 2·2^3 + 3·2^5 +.....+ (n-1)·2^(2n-3) + n · 2^(2n-1)
4Sn= 1·2^3 + 2·2^5 +.....+ (n-2)·2^(2n-3) + (n-1)·2^(2n-1) + n · 2^(2n+1)
两式相减：
-3Sn =1·2+2^3+2^5+....+2^(2n-3)+2^(2n-1)-n · 2^(2n+1)
=(-2/3)· [1-2^(2n)] - n · 2^(2n+1)
Sn = (n/3) · 2^(2n+1) - (2/9)· [2^(2n) -1 ]

2.
a3=7
a5+a7=26
∴a1=3，d = 2
an = a1+(n-1)d = 2n+1
Sn = na1+n(n-1)d/2=n²+2n

bn = 1/[(an)²-1] = (1/4)· [1/n(n+1)]
=(1/4)· [ 1/n - 1/(n+1)]
∴
b(n-1) = (1/4)· [ 1/(n-1) - 1/n]
.......
b3 = (1/4)· (1/3 - 1/4)
b2 = (1/4)· (1/2 - 1/3)
b1 = (1/4)· (1/1 - 1/2)
上述各式相加：
bn+b(n-1)+...+b3+b2+b1 = (1/4)· [1 - (1/n)]
=n/4(n+1)
因此，bn的前n项和为n/4(n+1)