Tính \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+\text{4}\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
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tính \(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2016}\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
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tính
A=\(\left(\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}\right)\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
Thuc hien phep tinh:
E=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
Tính M , biết :
\(M=1+\frac{1}{2}\times\left(1+2\right)+\frac{1}{3}\times\left(1+2+3\right)+\frac{1}{4}\times\left(1+2+3+4\right)+...+\frac{1}{2016}\times\left(1+2+3+4+...+2015+2016\right).\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+2016}\right)\).Tính A
thực hiện tính:
E= 1+\(\frac{1}{2}\) (1+2) + \(\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
Xét Sn = 1+2+3+4+...+n (1)
=> Sn= n+(n-1)+...+2+1 (2)
Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1
Lấy (1);(2) và chú ý trên ta có:
2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1) (vì n số hạng giống nhau)
=> Sn= n(n+1)/2 => Sn/n = (n+1)/2
=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n
P= 1+3/2+4/2+5/2+...+(n+1)/2
P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2
P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n2+3n
Bài toán chỉ đến S2016/2016 (tức n=2016)
Vậy S= 20162+3.2016=2016.(2016+3)=2016.2019=4070304
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E = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 + 3) + 1/4.(1 + 2 + 3 + 4) + ... + 2016.(1 + 2 + 3 + ... + 2016)
E = 1 + 1/2.(1 + 2).2:2 + 1/3.(1 + 3).3:2 + 1/4.(1 + 4).4:2 + ... + 2016.(1 + 2016).2016:2
E = 2/2 + 3/2 + 4/2 + 5/2 + ... + 2017/2
E = 2+3+4+5+...+2017/2
E = (2 + 2017).2016/2
E = 2019.1008
E = 2 035 152
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Xét Sn = 1+2+3+4+...+n \
=> Sn= n+(n-1)+...+2+1 \
Thấy 1+n = 2+(n-1) = 3+(n-2) = n-1+2=n+1
Lấy (1);(2) và chú ý trên ta có:
2.Sn = (n+1)+(n+1)+(n+1)+...+(n+1)=n(n+1) (vì n số hạng giống nhau)
=> Sn= n(n+1)/2 => Sn/n = (n+1)/2
=> P= 1+ S2/2 + S3/3 + S4/4 +...+ Sn/n
P= 1+3/2+4/2+5/2+...+(n+1)/2
P= 2(2+3+4+...+n+n+1) = 2(1+2+...n+n+1) - 2 = 2.S(n+1) - 2
P= 2.(n+1)(n+2)/2 -2 = (n+1)(n+2) -2 = n2+3n
Vậy S= 20162+3.2016=2016.(2016+3)=2016.2019:2=2035152
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Rút gọn các tổng sau:
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2015}\)
\(B=1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2016}\)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
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\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
RÚT GỌN : \(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)......\left(1-\frac{1}{2016^2}\right)\)