Tính giá trị biểu thức sau:
A=(1/2+1).(1/3+1).(1/4+1)........(1/99+1)
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Cách tìm BCNN:
(1 - 1/2)(1 - 1/3)(1 - 1/4) ... (1 - 1/99)
= 1/2*2/3*3/4*...*98/99
= 1/99
Ta có : (1 -1/2)(1-1/3)(1-1/4)..(1-1/99)
=1/2 .2/3.3/4....98/99
=1/99
A=2100-299-298-...-22-2-1
\(\Rightarrow\)2A=2101-2100-299-...-23-22-2
\(\Rightarrow\)2A+A=(2101-2100-299-...-23-22-2)+(2100-299-298-...-22-2-1)
\(\Rightarrow\)3A=2101+1
\(\Rightarrow\)A=\(\frac{2^{101}+1}{3}\)
Vậy A=\(\frac{2^{101}+1}{3}\).
Ta có : A = 2100 - 299 - 298 - ... - 22 - 2 - 1
=> 2A = 2101 - 2100 - 299 - ... - 23 - 22 - 2
Lấy A - 2A = (2100 - 299 - 298 - ... - 22 - 2 - 1) - (2101 - 2100 - 299 - ... - 23 - 22 - 2)
=> - A = 2100 + 2100 - 2101 - 1
=> - A = 2.2100 - 2101 - 1
=> - A = 2101 - 2101 - 1
=> - A = - 1
=> A = 1
#include <bits/stdc++.h>
using namespace std;
int n,i;
double s;
int main()
{
cin>>n;
s=1;
for (i=2; i<=n; i++)
{
if (i%2==0) s=s+1/(i*1.0);
else s=s-1/(i*1.0);
}
cout<<fixed<<setprecison(2)<<s;
return 0;
}
A=(1/2+1)*(1/3+1)*(1/4+1).....(1/99+1)
A=3/2*4/3*5/4.....100/99 (Thực hiện tính tổng trong mỗi ngoặc đơn)
A=(3*4*5...100)/(2*3*4...99)
A=100/2 (Rút gọn những thừa số giống nhau ở tử và mẫu)
A=50
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
Sửa:\(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)........\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}.\frac{4}{3}.............\frac{100}{99}=\frac{3.4....................100}{2.3.................99}=\frac{\left(3.4.......99\right).100}{2.\left(3.4...........99\right)}=\frac{100}{2}=50\)
Vậy A=50
A=\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right)..............\left(\frac{1}{99}+1\right)\)
=\(\frac{3}{2}.\frac{4}{3}.............\frac{100}{99}\)
=\(\frac{100}{2}\)=50