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\(1^2-2^2+3^2-4^2+5^2-6^2+........+99^2-100^2\)
\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right)+........+\left(99^2-100^2\right)\)
\(=\left(1+2\right)\left(1-2\right)+\left(3+4\right)\left(3-4\right)+\left(5+6\right)\left(5-6\right)........+\left(99+100\right)\left(99-100\right)\)
\(=-1\left(1+2\right)+-1\left(3+4\right)+-1\left(5+6\right)+........+-1\left(99+100\right)\)
\(=-1\left[\left(1+2\right)+\left(3+4\right)+\left(5+6\right)+........+\left(99+100\right)\right]\)
\(=-1\left(3+7+11+........+199\right)\)
\(=-1.\left\{\dfrac{\left(199+3\right).\left[\left(199-3\right):4+1\right]}{2}\right\}\)
\(=-1.\left[\dfrac{202.\left(196:4+1\right)}{2}\right]\)
\(=-1.\left[\dfrac{202.\left(49+1\right)}{2}\right]\)
\(=-1.\dfrac{202.50}{2}\)
\(=-1.\dfrac{10100}{2}\)
\(=-1.5050\)
\(=-5050\)
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Đề bài trên sai. Đề đúng: CM: \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{97}{98}.\dfrac{99}{100}>\dfrac{\sqrt{2}}{20}\).
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có
\(A=3\cdot11...111\cdot9\cdot11...111\)(100 số 1)
\(A=27\cdot\left(11...111\right)^2\)(100 số 1)
\(A=27\left(10^{99}+...+10^0\right)^2\)
\(A=27\left(\dfrac{10^{99+1}-1}{10-1}\right)^2\)
\(A=\dfrac{10^{200}-2\cdot10^{100}+1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3: a) Xét A=(1+1/2+1/3+....+1/98).2.3.4.5.....98
=(1+1/2+1/3+....+1/98).(9.11).2.3.4.....98
=(1+1/2+1/3+....+1/98).99.2.3.4....98⋮99
(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+4+...+100}\)
\(A=3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\right)\)
Đặt \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\), khi đó ta đc:
\(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+4+...+100}\)
Vì tổng số hạng bằng (số cuối + số đầu) . số số hạng : 2 nên ta có:
\(B=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+100\right).100:2}\)
\(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\)
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(B=2.\left(1-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
Ta có:
\(A=3.B\Rightarrow A=3.\frac{200}{101}=\frac{600}{101}\)
Vậy \(A=\frac{600}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em thử nhá, ko chắc đâu...
Gọi B là tổng các phần tử trong tập hợp A.
Thì \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
Xét dạng tổng quát \(\frac{1}{\sqrt{n}+\sqrt{n+1}}\left(n\in Z^+\right)=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra \(B=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=10-1=9\) là một số nguyên (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo công thức (n-1)n(n+1)=n\(^3\)-n
\(\Rightarrow\) n\(^3\)=n+(n-1)n(n-1)
Ta có :
\(A=1^3+2^3+.....+100^3\)
\(\Rightarrow1+2+1\cdot2\cdot3+3+2\cdot3\cdot4+100+99\cdot100\cdot101\)\(=\left(1+2+3+...+100\right)+\left(1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\right)\) =5050+101989800
=101994850