1.1+1/3+1/3的平方+.+1/3的n次方除以1+1/9+1/9的平方+.1/9的n-1次方.问当n→无穷大时,上述数列的极限?2.当n→无穷大时,a.n的平方+b.n+1/n+3=2 求:a+b+=?3.lim (1+q)(1+q的平方)(1+q的3次方)(1+q的2n次方)=?
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![1.1+1/3+1/3的平方+.+1/3的n次方除以1+1/9+1/9的平方+.1/9的n-1次方.问当n→无穷大时,上述数列的极限?2.当n→无穷大时,a.n的平方+b.n+1/n+3=2 求:a+b+=?3.lim (1+q)(1+q的平方)(1+q的3次方)(1+q的2n次方)=?](/uploads/image/z/5278325-5-5.jpg?t=1.1%2B1%2F3%2B1%2F3%E7%9A%84%E5%B9%B3%E6%96%B9%2B.%2B1%2F3%E7%9A%84n%E6%AC%A1%E6%96%B9%E9%99%A4%E4%BB%A51%2B1%2F9%2B1%2F9%E7%9A%84%E5%B9%B3%E6%96%B9%2B.1%2F9%E7%9A%84n-1%E6%AC%A1%E6%96%B9.%E9%97%AE%E5%BD%93n%E2%86%92%E6%97%A0%E7%A9%B7%E5%A4%A7%E6%97%B6%2C%E4%B8%8A%E8%BF%B0%E6%95%B0%E5%88%97%E7%9A%84%E6%9E%81%E9%99%90%3F2.%E5%BD%93n%E2%86%92%E6%97%A0%E7%A9%B7%E5%A4%A7%E6%97%B6%2Ca.n%E7%9A%84%E5%B9%B3%E6%96%B9%2Bb.n%2B1%2Fn%2B3%3D2+%E6%B1%82%EF%BC%9Aa%2Bb%2B%3D%3F3.lim+%281%2Bq%29%281%2Bq%E7%9A%84%E5%B9%B3%E6%96%B9%EF%BC%89%EF%BC%881%2Bq%E7%9A%843%E6%AC%A1%E6%96%B9%EF%BC%89%EF%BC%881%2Bq%E7%9A%842n%E6%AC%A1%E6%96%B9%EF%BC%89%3D%3F)
1.1+1/3+1/3的平方+.+1/3的n次方除以1+1/9+1/9的平方+.1/9的n-1次方.问当n→无穷大时,上述数列的极限?2.当n→无穷大时,a.n的平方+b.n+1/n+3=2 求:a+b+=?3.lim (1+q)(1+q的平方)(1+q的3次方)(1+q的2n次方)=?
1.1+1/3+1/3的平方+.+1/3的n次方除以1+1/9+1/9的平方+.1/9的n-1次方.问当n→无穷大时,上述数列的极限?
2.当n→无穷大时,a.n的平方+b.n+1/n+3=2 求:a+b+=?
3.lim (1+q)(1+q的平方)(1+q的3次方)(1+q的2n次方)=?
n→无穷大
就这些了,Crz
1.1+1/3+1/3的平方+.+1/3的n次方除以1+1/9+1/9的平方+.1/9的n-1次方.问当n→无穷大时,上述数列的极限?2.当n→无穷大时,a.n的平方+b.n+1/n+3=2 求:a+b+=?3.lim (1+q)(1+q的平方)(1+q的3次方)(1+q的2n次方)=?
1.由等比数列前n项和公式Sn=首项×(1-q的n次方)÷(1-q)得:
原式={[1-(1/3)的n次方]/[1-(1/9)的n次方]}×4
∵当n趋近于无穷大时,(1/3)的n次方和(1/9)的n次方都趋近于0
∴当n趋近于无穷大时,原式的极限为4.
厉害~~
我完全看不懂~~ ^_^
高考结束一年了,早在一年前也许会记得
1.将式子整理,分子:{3[1-(1/3)^n]}/2,{9[1-(1/9)^n-1]}/8
求极限得:分子:3/2,分母:9/8
为:4/3
累了待续!
...............晕了..............