这道小学数学题目1+3⼀2+5⼀2*2+7⼀2*2*2+9⼀2*2*2*2+11⼀2*2*2*2*2+13⼀2*2*2*2*2*2=?怎么计算?依次类推,如...

2024-12-25 02:03:15
推荐回答(5个)
回答1:

设一sn=1+(3/2+5/2^2+……+13/2^6) 乘以1/2
则二1/2sn=(1/2+3/2*2+……+11/2^6)+13/2^7
一二错位减一下 得到1/2sn=1+(1+1/2+1/2^2+…+1/2^5)-13/2^7
括号里是个等比数列用公式就算出来了 项数再加也一样做

回答2:

这个貌似是高一的内容了,可以看作是;(2*0+1)/2^0+(2*1+1)/2^1+...+(2*6)2^6=?
以此类推的话,就把每个变数该为N,就是等比、等差数列。小学生能解?难道现在的小学生都有如此水平了?我们还咋混啊~

回答3:

这个貌似是高一的内容了,可以看作是;(2*0+1)/2^0+(2*1+1)/2^1+...+(2*6)2^6=?
以此类推的话,就把每个变数该为N,就是等比、等差数列。

回答4:

设一sn=1+(3/2+5/2^2+……+13/2^6)乘以1/2
则二1/2sn=(1/2+3/2*2+……+11/2^6)+13/2^7
一二错位减一下 得到1/2sn=1+(1+1/2+1/2^2+…+1/2^5)-13/2^7

回答5:

S1=1
S2=1+3/2=5/2
...
Sn=(2n-3)*2^(n-2)+3/2

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