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Ta có : \(S=\frac{989898.89-898989.98}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=\frac{98\cdot10101\cdot89-89\cdot10101\cdot98}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=\frac{10101\cdot\left(98\cdot89-89\cdot98\right)}{2^3+3^4+4^5+....+2014^{2015}}\)
\(=\frac{10101\cdot0}{2^3+3^4+4^5+....+2014^{2015}}=0\)
Vậy \(S=0\)
\(S=\frac{989898.89-898989.98}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=\frac{98\cdot10101\cdot89-89\cdot10101\cdot98}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=\frac{10101\cdot\left(98\cdot89-89\cdot98\right)}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=\frac{10101\cdot0}{2^3+3^4+4^5+...+2014^{2015}}\)
\(=0\)
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S = ( 1 - \(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))....(1-\(\dfrac{1}{50^2}\))
S = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{50^2-1}{50^2}\)
Vì em lớp 6 nên phải làm thêm bước này nữa:
Ta có
n2 - 1 = n2 - n + n - 1 = (n2 - n) + (n - 1) = n(n-1) + (n-1) =(n-1)(n+1)
Áp dụng công thức vừa chứng minh trên vào tổng S ta có:
S = \(\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}\).\(\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}\)....\(\dfrac{\left(50-1\right)\left(50+1\right)}{50^2}\)
S = \(\dfrac{1.3}{2^2}\).\(\dfrac{2.4}{3^2}\)......\(\dfrac{49.51}{50^2}\)
S = \(\dfrac{\left(3.4.5.6....49\right)^2.1.2.50.51}{\left(3.4.5.6...49\right)^2.2.2.50.50}\)
S = \(\dfrac{1}{2}\) . \(\dfrac{51}{50}\)
S = \(\dfrac{51}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Vì S chia hết cho 2 nhưng không chia hết cho 4 => S không là SCP
Vậy S không là số chính phương
~HỌC TỐT NHÉ
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Ta có: D = 1 + 4 + 42 + 43 + .... + 4100
=> 4D = 4 + 42 + 43 + .... + 4101
=> 4D - D = 4101 - 1
=> 3D = 4101 - 1
=> D = 4101 - 1/3
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why help you ?? why ?
KILL, please!!! Help me