Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
Tính C=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}\)
Bạn nào giúp mik nhớ viết cả cách giải cho mik nhé!!!!!!!!!!
Tập hợp các giá trị nguyên dương của x thỏa mản :\(\left(\right)\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\left(\right)\times x<\frac{13}{7}\)có số phần tử là
thực hiện phép tính
A=\(\frac{3}{1\times5}+\frac{3}{5\times10}+....+\frac{3}{100\times105}\)
B=
\(\dfrac{5}{1\times3\times5}+\dfrac{5}{3\times5\times7}+...+\dfrac{5}{99\times101\times103}\)
Tính:\(\left(1-\frac{2}{2\times3}\right)\left(1-\frac{2}{3\times4}\right)\left(1-\frac{2}{4\times5}\right)....\left(1-\frac{2}{99\times100}\right)\)
Cho C = \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\). So sanh C với \(\frac{1}{2}\)
Cho C = \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\). so sánh c với \(\frac{1}{2}\)
\(\left(1-\frac{2}{2\times3}\right)\left(1-\frac{2}{3\times4}\right)\left(1-\frac{2}{4\times5}\right)...\left(1-\frac{2}{99\times100}\right)\)= ?
Chứng minh rằng: \(\frac{1\times2-1}{2!}+\frac{2\times3-1}{3!}+\frac{3\times4-1}{4!}+...+\frac{99\times100-1}{100!}<2\)
tính \(A=\frac{1}{2}\times\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{2015\times2017}\right)\)