Tính giá trị của ( 1 + 2/3 ) × (1 + 2/4 ) × (1 + 2/5)× · · · × (1 + 2/97) × (1 + 2/98)
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=5/3*6/4*7/5*.....*99/97*100/98
=5*6*7*...*99*100/3*4*5*...*97*98
=100/3
mk chưa chắc nhưng thấy sai mọi người cứ góp ý nha! nói thẳng ra luôn
Kết quả : ( 1 + 2/3 ) x ( 1 + 2/4 ) x ( 1 + 2/5 ) x ... x ( 1 + 2/97 ) x ( 1 + 2/98 )
= ( 3/3 + 2/3 ) x ( 4/4 + 2/4 ) x ( 5/5 + 2/5 ) x .... x ( 97/97 + 3/97 ) x ( 98/98 + 2/98 )
= 5/3 x 6/4 x 7/5 x ... x 99/97 x 100/98
= ( 5 x 6 x 7 x ...x 99 x 100 ) / ( 3 x 4 x 5 x ... x 97 x 98 )
= 99 x 100 / 3 x 4
= 33 x 25 / 1 x 1
= 825
Tham khảo nha !!!
Ai nhanh và đúng nhất mình k cho.
Mình đang cần gấp!!!!!
a)
C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.
b)
B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.
có 2 cách
C1 dùng xích ma \(\text{∑}^{97}_1\left(x.\left(x+1\right)\left(x+2\right)\right)=23527350\)
c2 dùng quy nạp \(\frac{97.98.99.100}{4}=23527350\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
(1+2+3+4+...+96+97+98+99):5
Đặt 1+2+3+4+...+96+97+98+99=A1+2+3+4+...+96+97+98+99=S
Số số hạng của S là:
(99−1):1+1=99(99-1):1+1=99
Tổng của S là:
(99+1).99:2=4950(99+1).99:2=4950
→(1+2+3+4+...+96+97+98+99):5→(1+2+3+4+...+96+97+98+99):5
=4950:5=990
S=(1+2+3+4+...+96+97+98+99):5
S=(99x(99+1):2):5
S=(99x100:2):5
S=(9900:2):5
S=4950:5
S=990
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\frac{4949}{9900}\)
\(A=\frac{4949}{19800}\)