1.设sn是等差数列{an}的前n项和,已知S6=36,Sn=324,若Sn-6=144(n》6),则n=
2.定义在R上的函数f(x)满足f(1/2+x)+f(1/2-x)=2.则f(1/8)+f(2/8)+f(3/8)+·····+f(1/8)=
人气:163 ℃ 时间:2019-12-16 21:37:08
解答
Sn-Sn-6=an+an-1+an-2+...+an-5=a6+(n-6)d+a5+(n-6)+...+a1+(n-6)d=S6+6(n-6)d所以d=24/(n-6)S6=3(a1+a6)=3(2a1+5d)所以2a1=12-5dSn=n/2*(a1+an)=(n/2)(2a1+(n-1)d)=(n/2)(12+(n-6)d)324=6n+(n-6)dn/2d=2(324-6n)/(n...
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