Tính hộ tui nha:>>
1/1.2.3+1/2.3.4+1/3.4.5+...+1/27.28.29+1/28.29.30
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Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
a) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{27.28}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\frac{405}{812}=\frac{405}{1624}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}=\frac{405}{1624}\)
b) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+....+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}\cdot\frac{1353}{8120}=\frac{451}{8120}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}=\frac{451}{8120}\)
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
gọi A=................................
=>2A=2/1.2.3+2/2.3.4+.....+2/8.9.10
2A=1/1.2-1/2.3+1/2.3-...+1/8.9-1/9.10
2A=1/1.2-1/9.10=22/45 =>A=11/45
Đặt S = 1/1.2.3 - 1/2.3.4 - 1/3.4.5 - ...- 1/97.98.99
S x 2 = 2/1.2.3 - 2/2.3.4 - 2/3.4.5 - ...- 2/97.98.99
= (1/1.2 -1/2.3) - (1/2.3 - 1/3.4 ) - (1/3.4 - 1/4.5) - ...- (1/97.98 - 1/98.99)
= 1/1.2 - 1/2.3 - 1/2.3 + 1/3.4 - 1/3.4 + 1/4.5 - ....- 1/97.98 + 1/98.99
= 1/2 -1/3 + 1/98.99
= 1618/9072 => S = 1618/9072 : 2 = 809/9072
\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1353}{8120}\)