0001以下、5ちゃんねるからVIPがお送りします
2022/09/16(金) 18:02:36.920ID:Lkgi9jgq0b_n= b[n]=cos(π/(2^n))+cos(π/(2^(n-1)))
a[n]=cos(π/(2^n))-cos(π/(2^(n-1)))
=-2sin{{(π/(2^n))+(π/(2^(n-1)))}/2}*sin{{(π/(2^n))-(π/(2^(n-1)))}/2}
=2sin(3π/(2^(n+1)))/*sin(π/(2^(n+1)))
lim[n→∞]4^n・a_n=2・(1/2)^2・(1/3π^2)=1/6π^2
になるぽいわ！！
https://i.imgur.com/p6z2BTA.jpg
