Đề bài: Chứng minh rằng:$C_{n}^{0}-\frac{1}{3}C_{n}^{1}+\frac{1}{5}C_{n}^{2}+…+\frac{(-1)^{n}}{2n+1}C_{n}^{n}\geq \sqrt{\frac{3n+1}{4n^{2}+4n+1}}$

Lời giải

Đề bài:
Chứng minh rằng:$C_{n}^{0}-\frac{1}{3}C_{n}^{1}+\frac{1}{5}C_{n}^{2}+…+\frac{(-1)^{n}}{2n+1}C_{n}^{n}\geq \sqrt{\frac{3n+1}{4n^{2}+4n+1}}$
Lời giải

Xét: $\int\limits^{1}_{0}(1-x^{2})^{n}dx,n \in N^{*}$
* Đặt: $x=\sin t \Rightarrow  dx=\cos tdt$
  $\Rightarrow \int\limits^{1}_{0}(1-x^{2})^{n}dx=\int\limits^{\frac{\pi}{2}}_{0} \cos ^{2n+1}t.dt=I_{2n+1}$
* Đặt: $u=\cos ^{2n}t \Rightarrow du=-2n \sin t.\cos t ^{2n-1}dt$
$dv=\cos tdt \Rightarrow v=\sin t \Rightarrow I_{2n+1}=[\cos ^{2n}t.\sin t]^{\frac{\pi}{2}}_{0}+2n\int\limits^{\frac{\pi}{2}}_{0}\sin ^{2}t \cos ^{2n-1}t.dt$
$=2n \int\limits^{\frac{\pi}{2}}_{0} (\cos ^{2n-1}t-\cos ^{2n+1}t).dt=2nI_{2n-1}-2nI_{2n+1}$
$\Rightarrow \frac{I_{2n+1}}{I_{2n-1}}=\frac{2n}{2n+1}$
Suy ra: $\frac{I_{3}}{I_{1}}.\frac{I_{5}}{I_{3}}…\frac{I_{2n}}{I_{2n+1}}=\frac{2}{3}.\frac{4}{5}…\frac{2n}{2n+1}$
$\Rightarrow I_{2n+1}=\frac{2}{3}.\frac{4}{5}…\frac{2n}{2n+1}.I_{1}$
Mà: $I_{1}=\int\limits^{1}_{0} dx=1$
$\Rightarrow I_{2n+1}=\frac{2}{3}.\frac{4}{5}…\frac{2n}{2n+1} (1)$
và: $ I_{2n+1}=\int\limits^{1}_{0} (1-x^{2})^{n}dx=\int\limits^{1}_{0}\sum\limits_{k=0}^n C^{k}_{n}.(-1)^{k}.x^{2k}dx$
$=\sum\limits_{k=0}^n C^{k}_{n}.\frac{(-1)^{k}}{2k+1} (2)$
Từ $(1)$ và $(2)$ $\Rightarrow C_{n}^{0}-\frac{1}{3}C_{n}^{1}+\frac{1}{5}C_{n}^{2}+…+\frac{(-1)^{n}}{2n+1}C_{n}^{n}=\frac{2}{1}.\frac{4}{3}…\frac{2n}{2n-1}\frac{1}{2n+1}$
$\geq \sqrt{3n+1}.\frac{1}{2n+1}= \sqrt{\frac{3n+1}{4n^{2}+4n+1}}$.Đúng.
(Theo nguyên lý quy nạp:ta chứng minh bài toán nhỏ:$ \frac{1}{2}.\frac{3}{4}…\frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$
*$n=1$: BĐT luôn đúng.
*$n=k$: Giả sử BĐT đúng,tức là:
$\frac{1}{2}.\frac{3}{4}…\frac{2k-1}{2k}.\frac{2k+1}{2k+2}\leq \frac{1}{\sqrt{3k+1}}.\frac{2k+1}{2k+2}(3)$
$(\frac{1}{\sqrt{3k+1}}.\frac{2k+1}{2k+2})^{2}=\frac{(2k+1)^{2}}{(3k+1)(4k^{2}+8k+4)}=\frac{(2k+1)^{3}}{12k^{3}+28k^{2}+20k+4}$
$=\frac{(2k+1)^{2}}{(12k^{3}+28k^{2}+19k+4)+k}=\frac{(2k+1)^{2}}{(2k+1)^{2}(3k+4)+k}$
$Từ $(3)$ và $(4)$ suy ra: $\frac{1}{2}.\frac{3}{4}…\frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$)
$\Rightarrow$ (ĐPCM)

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