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a) Số hạng trong tổng có dạng n.(n+1)(n+2)
nhận xét: n(n+1)(n+2)(n+3) - (n-1).n(n+1)(n+2) = 4.n(n+1)(n+2). Tính A
4.A = 2.3.4.(5-1) + 3.4.5.(6-2) + ...+ 20.21.22.(23 - 19)
4.A = 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ...+ 20.21.22.23 - 19.20.21.22
4.A = (2.3.4.5 + 3.4.5.6 + ...+ 20.21.22.23) - (1.2.3.4 + 2.3.4.5 + ...+ 19.20.21.22)
4.A = 20.21.22.23 - 1.2.3.4 = 212 496 => A = 53 124
b) Em xem lại : dạng nào đã hỏi rồi , em nên tự làm
\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)
Bài này bạn làm theo công thức:1/axa+1=1/a-1/a-1
Ta có:
A=1/5-1/6+1/6-1/7+1/7-1/8+...+1/99-1/100
A=1/5-1/100
A=19/100
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)\(\frac{1}{25}\)
\(A=\frac{1}{5}-\frac{1}{25}\)
\(A=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(A=\frac{1}{5}-\frac{1}{25}\)
\(A=\frac{4}{25}\)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
Ez :))
~~~HD~~~
Đặt: A=5.6+6.7+..........+97.98
=> 3A=5.6(7-4)+6.7(8-5)+.......+97.98(99-96)
=> 3A=97.98.99-4.5.6=941094-120=940984
1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25
=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25
=> Kết quả là: 1/5 - 1/25 = 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101
=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101
=> kết quả là 2/1 - 2/101 =200/101
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
=\(\frac{1}{5}-\frac{1}{25}\)
=\(\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
=\(2.\frac{100}{101}\)
=\(\frac{200}{101}\)
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{21}-\frac{1}{22}\)
\(A=\frac{1}{5}-\frac{1}{22}\)
\(A=\frac{17}{110}\)
\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+...+\(\frac{1}{21.22}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+...+\(\frac{1}{21}\)-\(\frac{1}{22}\)
=\(\frac{1}{5}\)-\(\frac{1}{22}\)
=\(\frac{17}{110}\)
A.3=5.6.3+6.7.3+...+30.31.3
A.3=5.6.(7-4)+6.7.(8-5)+...+30.31.(32-29)
A.3=5.6.7-4.5.6+6.7.8-5.6.7+...+30.31.32-29.30.31
A.3=(5.6.7-5.6.7)+...+(29.30.31-29.30.31)+(30.31.32-4.5.6)
A.3=0+...+0+30.31.32-4.5.6
A.3=30.31.32-4.5.6
A=30.31.32-4.5.6 /3
A=(29760-120)/3
A=29460/3
A=9880
vậy A là 9880
lưa ý dấu này/ nghĩa là chia
3.A = 5.6.(7-4) + 6.7.(8-5) + ....+30.31.(32- 29)
3.A = 5.6.7 - 4.5.6 + 6.7.8 - 5.6.7 + ...+ 30.31.32 - 29.30.31
3.A = (5.6.7 + 6.7.8 + ...+ 30.31.32) - (4.5.6 + 5.6.7 + ...+ 29.30.31)
3.A = 30.31.32 - 4.5.6 = 29 640 => A = 9 880