(1+1/2).(1+1/3).(1+1/4)......(1+1/300)
Những câu hỏi liên quan
CMR: 1/151+1/152+1/153+…+1/300=1-1/2+1/3-1/4+…+1/299-1/300
CMR:S=1/2+1/3+1/4+…1/2013
so sánh :
a.3^300 +4^300 và 3.24^100
b.(20^2006 + 11^2006)^2007 và (20^2007 +11^2007)^2006
c.(1/2^2-1).(1/3^2-1).(1/4^2-1)..........(1/1000^2-1) và -1/2
so sanh A= 1/2^2 + 1/ 3^2 +1/4^2+...+ 1/300^2 voi 3/4
1/2+1/3+1/4+...+1/300
___________________
2999/1+2998/2+2997/3+...+1/2999
Đề là 1/3000 nhé ~
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\left(\frac{2998}{2}+1\right)+\left(\frac{2997}{3}+1\right)+...+\left(\frac{1}{2999}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\frac{3000}{2}+\frac{3000}{3}+....+\frac{3000}{2999}+\frac{3000}{3000}}\)
\(=\frac{1}{3000}\)
Tính hợp lý:
a. S= 1+ 1/2 + 1/2^2 + 1/2^3 +...+ 1/2^100
b. D = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +...+ 1/3^300
a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2S=2+1+...+\frac{1}{2^{99}}\)
\(2S-S=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(S=2-\frac{1}{2^{100}}\)
phần b tương tự
Đúng 0
Bình luận (0)
a. S=1+1/2+1/2^2+1/2^3+...+1/2^100
2S=2+1+1/2+1/2^2+...+1/2^99
2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)
S=2-1/2^100
S=2^101-1/2^100
Đúng 0
Bình luận (0)
b. D=1/3+1/3^2+1/3^3+1/3^4+...+1/3^300
3D=1+1/3+1/3^2+1/3^3+...+1/3^299
3D-D=(1+1/3+1/3^2+1/3^3+...+1/3^299)-(1/3+1/3^2+1/3^3+1/3^4+...+1/3^300)
2D=1-1/3^300
D=1-1/3^300/2
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
So sánh
a) 3^300+4^300 và 3.24^100
b) 2^23+1/2^28+1 và 2^25+1/2^24+1
tịnh 1+20+300+4000+50000+600000+7000000+80000000+900000000+1/2-1/3*1/4*1/5
giỏi đấy!
2 2 2=6
3 3 3=6
6 6 6=6
hãy điền dấu thích hợp
Đúng 0
Bình luận (0)
Tính tích P = ( 1- 1/2). (1- 1/3). (1- 1/4) ... ( 1 -1/99)
Chứng tỏ rằng : 1/101 +1/102+ ... +1/299 + 1/300 > 2/3
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{98}{99}=\dfrac{1.2.3...98}{2.3.4...99}=\dfrac{1}{99}\)
Đúng 0
Bình luận (0)
1+2+3+4+..............+n(300) hoi n=?
biet rang [(1+n)].n:2=300
Ta có n.(n+1) =2.300
n.(n+1)=600
n.(n+1)=23.3.52
n.(n+1)=24.25
vậy n=24