已知数列an=10-n,求数列{|an|}的前n项和Sn设f(x)=4^x/4^x+2,求f(1/2011)+f(2/2011)+.+f(2010/2011)的值Sn=1²-2²+3²-4²+5²-6²+.+99²-100²,求Sn
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![已知数列an=10-n,求数列{|an|}的前n项和Sn设f(x)=4^x/4^x+2,求f(1/2011)+f(2/2011)+.+f(2010/2011)的值Sn=1²-2²+3²-4²+5²-6²+.+99²-100²,求Sn](/uploads/image/z/7646418-18-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%3D10-n%2C%E6%B1%82%E6%95%B0%E5%88%97%7B%7Can%7C%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E8%AE%BEf%EF%BC%88x%EF%BC%89%3D4%5Ex%2F4%5Ex%2B2%2C%E6%B1%82f%EF%BC%881%2F2011%29%2Bf%282%2F2011%29%2B.%2Bf%282010%2F2011%29%E7%9A%84%E5%80%BCSn%3D1%26%23178%3B-2%26%23178%3B%2B3%26%23178%3B-4%26%23178%3B%2B5%26%23178%3B-6%26%23178%3B%2B.%2B99%26%23178%3B-100%26%23178%3B%2C%E6%B1%82Sn)
已知数列an=10-n,求数列{|an|}的前n项和Sn设f(x)=4^x/4^x+2,求f(1/2011)+f(2/2011)+.+f(2010/2011)的值Sn=1²-2²+3²-4²+5²-6²+.+99²-100²,求Sn
已知数列an=10-n,求数列{|an|}的前n项和Sn
设f(x)=4^x/4^x+2,求f(1/2011)+f(2/2011)+.+f(2010/2011)的值
Sn=1²-2²+3²-4²+5²-6²+.+99²-100²,求Sn
已知数列an=10-n,求数列{|an|}的前n项和Sn设f(x)=4^x/4^x+2,求f(1/2011)+f(2/2011)+.+f(2010/2011)的值Sn=1²-2²+3²-4²+5²-6²+.+99²-100²,求Sn
第一题,n=10时,Sn=-(a1+a2+a3+……)+2(a1+a2+……+a9)=-(9+10-n)n/2+90=(n^2-19n)/2+90.
第二题实在是看不清楚你是怎么样写的题目
第三题:1²-2²+3²-4²+5²-6²+.+99²-100²=(1-2)(1+2)+(3-4)(3+4)+……+(99-100)(99+100)=-(1+2+3+4+……+100)我相信你应该会算了
(1) sn=n(19-n)/2
(2) f(x)+f(1-x)=1,f(1/2011)+f(2/2011)+.....+f(2010/2011)=1005
(3) Sn=-(1+2)-(3+4)-.....-(99+100)=-50(1+100)=-5050
Sn=10n-(1+2+3+……+n)=10n-(1+n)n/2=-0.5n^2+9.5n(等差数列简单应用) 1005(倒序相加,同分,除以2) -5050(平方差公式)