【高一数学】如函数f(x)=1/[(2^x)+根号2] 求T=f(-5)+f(-4)+f(-3)+...+f(0)+..
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/09 02:35:56
【高一数学】如函数f(x)=1/[(2^x)+根号2] 求T=f(-5)+f(-4)+f(-3)+...+f(0)+...+f(5)+f(6)
有没有简便方法?
有没有简便方法?
/>f(x)=1/(2^x+√2)
f(-4)+f(-3)+……+f(5)
=1/[2^(-4)+√2]+1/[2^(-3)+√2]+1/[2^(-2)+√2]+1/[2^(-1)+√2]+1/(2^0+√2)+1/(2^1+√2)+1/(2^2+√2)+
+1/(2^3+√2)+1/(2^4+√2)+1/(2^5+√2)
=16/(1+16√2)+8/(1+8√2)+4/(1+4√2)+2/(1+2√2)+1/(1+√2)+1/(2+√2)+1/(4+√2)++1/(8+√2)+
+1/(16+√2)+1/(32+√2)
=(16-256√2)/(-511)+(8-64√2)/(-127)+(4-16√2)/(-31)+(2-4√2)/(-7)+(1-√2)/(-1)+(2-√2)/2+
+(4-√2)/14+(8-√2)/62+(16-√2)/254+(32-√2)/1022
=(256/511)√2-16/511+(64/127)√2-8/127+(16/31)√2-4/31+(4/7)√2-2/7+√2-1+1-(√2)/2+2/7-(√2)/14+
+4/31-(√2)/62+8/127-(√2)/254+16/511-(√2)/1022
=(256/511+64/127+16/31+4/7+1/2-1/14-1/62-1/254-1/1022)(√2)-16/511-8/127-4/31-
-2/7+2/7+4/31+8/127+16/511
=[1/2+(8/14-1/14)+(32/62-1/62)+(128/254-1/254)+(512/1022-1/1022)](√2)
=(1/2+1/2+1/2+1/2+1/2)(√2)
=(5/2)√2
再问: 这式子太烦了有简便方法嘛?
再答: 这就是最简单易懂的笨办法嘛 就是详细了点 不是烦吧 谢谢
f(-4)+f(-3)+……+f(5)
=1/[2^(-4)+√2]+1/[2^(-3)+√2]+1/[2^(-2)+√2]+1/[2^(-1)+√2]+1/(2^0+√2)+1/(2^1+√2)+1/(2^2+√2)+
+1/(2^3+√2)+1/(2^4+√2)+1/(2^5+√2)
=16/(1+16√2)+8/(1+8√2)+4/(1+4√2)+2/(1+2√2)+1/(1+√2)+1/(2+√2)+1/(4+√2)++1/(8+√2)+
+1/(16+√2)+1/(32+√2)
=(16-256√2)/(-511)+(8-64√2)/(-127)+(4-16√2)/(-31)+(2-4√2)/(-7)+(1-√2)/(-1)+(2-√2)/2+
+(4-√2)/14+(8-√2)/62+(16-√2)/254+(32-√2)/1022
=(256/511)√2-16/511+(64/127)√2-8/127+(16/31)√2-4/31+(4/7)√2-2/7+√2-1+1-(√2)/2+2/7-(√2)/14+
+4/31-(√2)/62+8/127-(√2)/254+16/511-(√2)/1022
=(256/511+64/127+16/31+4/7+1/2-1/14-1/62-1/254-1/1022)(√2)-16/511-8/127-4/31-
-2/7+2/7+4/31+8/127+16/511
=[1/2+(8/14-1/14)+(32/62-1/62)+(128/254-1/254)+(512/1022-1/1022)](√2)
=(1/2+1/2+1/2+1/2+1/2)(√2)
=(5/2)√2
再问: 这式子太烦了有简便方法嘛?
再答: 这就是最简单易懂的笨办法嘛 就是详细了点 不是烦吧 谢谢
高一数学 已知2f(x)+f(-x)=2x-3 求函数f(x)的值
【高一数学】函数f(x)满足f(x)=-f(x+3/2)且f(-2)=f(-1)=-1,f(0)=2,则f(1)+f(2
高一数学数学换元法已知:f(根号x)+1=x+2根号x求f(x)
一道高一函数数学题f(x)+2f(x+1/1-x)=3x,求f(x)
高一数学函数f(x)=【x^3+(x+1)^2】/x^2+1,x∈【-2,2】求f(x)最小值+f(x)最大值
已知函数F(X)=1+X平方分之X平方,求F(1)+F(2)+F(二分之一)+F(3)+F(三分之一)+F(4)+F(四
030高一数学已知f(3+x)+2f(1-x)=x^2问题补充:求f(x)
高一数学f(x)=(4^x-1)/(2^(x+1))-2x+1,已知f(m)=根号2,求f(-m)
高一数学 已知奇函数y=f(x) (x∈R) 满足f(x-2) = -f(x), 则f(1)+f(2)+f(3)=?
设F(x)是f(x)的一个原函数,f(x)F(x)=x+x^3,且F(0)=1/根号2,F(x)> 0,求f(x)
高数:设可导函数f(x)满足f(x)cosx+2∫(0~x)f(t)sintdt=x+1,求f(x)
已知函数f(x)=3x平方—5x+2,求f(负根号2),f(-a),f(a+3),f(a)+f(3)的值