Chứng Minh Rằng
1/2! + 2/3! + 3/4!+...+n/(n+1)! < 1
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Ta cần chứng minh:\(1^3+2^3+3^3+....+n^3=\left[\frac{n\left(n+1\right)}{2}\right]^2\)
Với \(n=1\Rightarrow1=1\)(đúng)
Giả sử bài toán đúng với \(n=k\left(n\inℕ^∗\right)\) thì ta có:
\(1+2^3+3^3+...+k^3=\left[\frac{n\left(n+1\right)}{2}\right]^2\left(1\right)\)
Ta cần chứng minh đề bài đúng với \(n=k+1\) tức là:
\(1^3+2^3+3^3+....+n^3=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\left(2\right)\)
Đặt \(A_{k+1}=1^3+2^3+...+\left(k+1\right)^3\)
\(=\left(\frac{k\left(k+1\right)}{2}\right)^2+\left(k+1\right)^3\) [theo (1)]
\(=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)
\(\Rightarrow\left(2\right)\) đúng
\(\Rightarrow\left(1\right)\) đúng.
Mà \(\left[\frac{n\left(n+1\right)}{2}\right]^2=\frac{n^2\cdot\left(n+1\right)^2}{4}\)
\(\Rightarrow1^3+2^3+...+n^3=\frac{n^2\cdot\left(n+1\right)^2}{4}\left(đpcm\right)\)
-Với n=1, ta thấy bthức đúng.
-Với n=k, có: \(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2k-1}{4+\left(2k-1\right)^4}=\frac{k^2}{4k^2+1}=\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\)
-Giả sử bthức đúng với n=k+1, có:
\(\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4\left(k+1\right)^2+1}\right)-\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\right)\)
\(=\frac{1}{4}\left(\frac{1}{4k^2+1}-\frac{1}{4\left(k+1\right)^2+1}\right)\)
\(=\frac{2k+1}{\left(4k^2+1\right)\left(4\left(k+1\right)^2+1\right)}=\frac{2k+1}{4+\left(2k+1\right)^4}\)
Vậy ta có đpcm.
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{\left(n+1\right)-1}{\left(n+1\right)!}\)
\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+...+\dfrac{\left(n+1\right)}{\left(n+1\right)!}-\dfrac{1}{\left(n+1\right)!}\)
\(=1-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}-\dfrac{1}{\left(n+1\right)!}\)
( Vì \(\dfrac{3}{3!}=\dfrac{1}{2!};\dfrac{4}{4!}=\dfrac{1}{3!};...;\dfrac{n+1}{\left(n+1\right)!}=\dfrac{1}{n!}\))
\(=1-\dfrac{1}{\left(n+1\right)!}< 1\)
Đặt \(S\left(n\right)=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)
Ta có \(S\left(1\right)=\dfrac{1}{2!}=\dfrac{1}{2}=1-\dfrac{1}{2!}\)
\(S\left(2\right)=S\left(1\right)+\dfrac{2}{3!}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}=1-\dfrac{1}{3!}\)
\(S\left(3\right)=S\left(2\right)+\dfrac{3}{4!}=\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{23}{24}=1-\dfrac{1}{4!}\)
Từ đây, ta có \(S\left(n\right)=1-\dfrac{1}{\left(n+1\right)!}\) và hiển nhiên \(S\left(n\right)< 1\) do \(\dfrac{1}{\left(n+1\right)!}>0\)
Vậy ta có đpcm