Tính:
\(C=\) \(\frac{1.1!}{1!.2!}+\frac{2.2!}{2!.3!}+\frac{3.3!}{3!.4!}+......+\frac{100.100!}{100!.101!}\)
chung to :C = \(\frac{1}{1.1!}+\frac{1}{2.2!}+\frac{1}{3.3!}+...+\frac{1}{2019.2019!}< \frac{3}{2}\)
Thấy : \(\frac{1}{1.1!}=\frac{1}{1}\)
\(\frac{1}{2.2!}=\frac{1}{4}\)
\(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )
...
\(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)
Cộng từng vế có :
\(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)
Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)
tìm giá trị của C , D. Biết :
C=100.100+1.1/100.1+99.99+2.2/99.2+...........+51.51+50.50/51/1/2+1/2+........+1/100+1/101D = 1+1/2.(1+2+3+4)+........+1/2016.(1+2+3+4)
2.Tính các tổng sau:
a)A=4.5.6+5.6.7+6.7.8+.....+49.50.51
b)B=1002+1012+1022+....+2002
3.Tính S=1+1.1!+2.2!+3.3!+....+100.100!
4.Tính S=9+99+999+9999+...+999....9(100 số 9)
5.Tìm số tự nhiên n lớn nhất để tích các số từ 1 đến 100 chia hết cho5
đề câu số 5 là chia hết cho \(5^n\)chứ ko phải là 5 đâu bạn
chứng tỏ \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + .........+ \(\frac{1}{100.100}\) < 1
Ta có : \(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4.4}< \frac{1}{3.4}\)
...................
\(\frac{1}{100.100}< \frac{1}{99.100}\)
Suy Ra : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+......+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Ta có : \(\frac{1}{2.2}\)\(< \frac{1}{1.2}\)
\(\frac{1}{3.3}\)\(< \frac{1}{2.3}\)
\(\frac{1}{4.4}\)\(< \frac{1}{3.4}\)
...... .... ......
\(\frac{1}{100.100}\)\(< \frac{1}{99.100}\)
\(\Rightarrow\)\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ..... + \(\frac{1}{100.100}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ..... + \(\frac{1}{99.100}\)
\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ .... + \(\frac{1}{100.100}\)< \(1-\frac{1}{100}=\frac{99}{100}< 1\)
1/2.2 < 1/1.2
1/3.3 < 1/2.3
1/4.4 < 1/3.4
1/100.100 < 1/ 99.100
Nên 1/2.2 + 1/3.3 +1/4.4 + .... +1/100.100 < 1/1.1 +1/2.3+1/3.4 +......+ 1/99.100
1/2.2 + 1/3.3+.... 1/100.100 < 1 - 1/100 = 99/100 < 1
ta còn có 1 cách làm ngắn gọn hơn
Giup mk tich cho
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+...........+\(\frac{1}{100.100}\)so sanh voi 1
Đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}\)
Mà \(\frac{99}{100}< 1\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\)
tính tổng D=1+1.1!+2.2!+3.3!+...+100.100!
Cho A= ( \(\frac{1}{2.2}\)-1).( \(\frac{1}{3.3}\)-1).( \(\frac{1}{4.4}\)-1)...( \(\frac{1}{100.100}\)-1)
So sánh A với -\(\frac{1}{2}\)
tính s
s1=3+7+11+...+2015
s2=1+2+4+5+7+8+...+100+101
s3=1.1+2.2=3.3+...+99.99
S1 = 3 + 7 + 11 + .... + 2015
SSH : ( 2015 - 3 ) : 4 + 1 = 504
Tổng : ( 2015 + 3 ) . 504 : 2 = 508536
S1 =1017072
S2=5151
S3(VIẾT SAI ĐẦU BÀI)
So sánh
A=1.1!+2.2!+3.3!+...+100.100!/1.199+2.197+3.195+...+100.1
B=99!/3
(! là giai thừa nhé.vd:2=1×2,3=1×2×3,4=1×2×3×4...)