Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
[(1+100)*100]/2= 5050
SỐ ĐẦU + SỐ CUỐI NHÂN CHO SỐ SỐ HẠNG RỒI CHIA TẤT CẢ CHO 2
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
1+2+3+4+5+6+7+8+9+..................+100=(1+99)+(2+98)+(3+97)+(4+96)+.......+100=100+100+100+.....+100=5000
chuc may man hihi
1+2+3+....+100
Từ 1 đến 100 có 100 số hạng
\(\Rightarrow1+2+3+...+100\)
\(=\frac{\left(100+1\right).100}{2}=5050\)
D=1/22+1/32+1/42+1/52+....+1/102+1/112
1/22<1/1x2 ; 1/32<1/2x3;...
=)D<1/1x2+1/2x3+1/3x4+1/4x5+...+1/9x10+1/10x11
D<1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11
D<1-1/11
D<10/11
Ta có: A = (1 - 2 ) + ( 3 -4 ) + ( 5 -6 ) + ...... + (99 - 100)
= -1 + -1 + ........ + -1
= -50
A=1-2+3-4+...+99-100( có 100 số,100 chia hết cho 2)
A=(1-2)+(3-4)+...+(99-100)( có 100:2=50 nhóm)
A= -1 + (-1) +...+ (-1)(50 số -1)
A= -1.50
A= -50
Câu B sai đề
\(1-2-3+4+5-6-7+8+...+97-98-99+100+101\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(97-98-99+100\right)+101\)
\(=0+0+...+0+101\)
\(=101\)
(100/7+100/9+100/11):(1/7+1/9+1/11)
=100.(1/7+1/9+1/11):(1/7+1/9+1/11)
=(100:1)(1/7+1/9+1/11)
=100.239/693
=23900/693
\(\left(\frac{100}{7}+\frac{100}{9}+\frac{100}{11}\right)\div\left(\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)\)
= \(100\times\left(\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)\div\left(\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)\)
= \(100\)