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Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
Lời giải:
$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$
$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$
\(Vì\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};.....;\dfrac{1}{n^2}< \dfrac{1}{(n-1).n}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{\left(n-1\right).1}< 1\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\left(đpcm\right)\)
vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\)
ĐặtA= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}\)
Do \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.............
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta suy ra : A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{\left(n-1\right)n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-....-\dfrac{1}{\left(n-1\right)}+\dfrac{1}{n-1}-\dfrac{1}{n}\)
=\(1-\dfrac{1}{n}\)
Mà 1-\(\dfrac{1}{n}\)<1
=> A<1 (đpcm)
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)
\(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)
Do đó:
\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+...+59}=\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{59}-\dfrac{2}{60}\)
\(=\dfrac{2}{3}-\dfrac{2}{60}< \dfrac{2}{3}\) (đpcm)
Đặt :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+................+\dfrac{1}{2^n}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{n-1}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{n-1}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^n}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^n}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...............+\dfrac{1}{2^n}< 1\rightarrowđpcm\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
................
\(\dfrac{1}{2^n}< \dfrac{1}{n.\left(n-1\right)}\)
\(\)- > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2^n}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)= \(1-\dfrac{1}{n}< 1\left(ĐPCM\right)\)