chứng minh \(1^2+2^2+...+n^2=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n\)
Chứng minh: \(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}.\dfrac{4^3+1}{4^3-1}....\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}\)
\(B=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{n!}< 1\)
\(C=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+....+\dfrac{n-1}{n!}< 1\)
D=\(\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)....\left(1-\dfrac{2}{n\left(n+1\right)}\right)>\dfrac{1}{3}\)
Bằng phương pháp quy nạp, chứng minh các đẳng thức sau với \(n\in N^{\circledast}\)
a) \(A_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)}\)
b) \(B_n=1+3+6+10+...+\dfrac{n\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)\left(n+2\right)}{6}\)
c) \(S_n=\sin x+\sin2x+\sin3x+...+\sin nx=\dfrac{\sin\dfrac{nx}{2}\sin\dfrac{\left(n+1\right)x}{2}}{\sin\dfrac{x}{2}}\)
b)
Với n = 1.
\(VT=B_n=1;VP=\dfrac{1\left(1+1\right)\left(1+2\right)}{6}=1\).
Vậy với n = 1 điều cần chứng minh đúng.
Giả sử nó đúng với n = k.
Nghĩa là: \(B_k=\dfrac{k\left(k+1\right)\left(k+2\right)}{6}\).
Ta sẽ chứng minh nó đúng với \(n=k+1\).
Nghĩa là:
\(B_{k+1}=\dfrac{\left(k+1\right)\left(k+1+1\right)\left(k+1+2\right)}{6}\)\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\).
Thật vậy:
\(B_{k+1}=B_k+\dfrac{\left(k+1\right)\left(k+2\right)}{2}\)\(=\dfrac{k\left(k+1\right)\left(k+2\right)}{6}+\dfrac{\left(k+1\right)\left(k+2\right)}{2}\)\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\).
Vậy điều cần chứng minh đúng với mọi n.
c)
Với \(n=1\)
\(VT=S_n=sinx\); \(VP=\dfrac{sin\dfrac{x}{2}sin\dfrac{2}{2}x}{sin\dfrac{x}{2}}=sinx\)
Vậy điều cần chứng minh đúng với \(n=1\).
Giả sử điều cần chứng minh đúng với \(n=k\).
Nghĩa là: \(S_k=\dfrac{sin\dfrac{kx}{2}sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}\).
Ta cần chứng minh nó đúng với \(n=k+1\):
Nghĩa là: \(S_{k+1}=\dfrac{sin\dfrac{\left(k+1\right)x}{2}sin\dfrac{\left(k+2\right)x}{2}}{sin\dfrac{x}{2}}\).
Thật vậy từ giả thiết quy nạp ta có:
\(S_{k+1}-S_k\)\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}sin\dfrac{\left(k+2\right)x}{2}}{sin\dfrac{x}{2}}-\dfrac{sin\dfrac{kx}{2}sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}\)
\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}.\left[sin\dfrac{\left(k+2\right)x}{2}-sin\dfrac{kx}{2}\right]\)
\(=\dfrac{sin\dfrac{\left(k+1\right)x}{2}}{sin\dfrac{x}{2}}.2cos\dfrac{\left(k+1\right)x}{2}sim\dfrac{x}{2}\)\(=2sin\dfrac{\left(k+1\right)x}{2}cos\dfrac{\left(k+1\right)x}{2}=2sin\left(k+1\right)x\).
Vì vậy \(S_{k+1}=S_k+sin\left(k+1\right)x\).
Vậy điều cần chứng minh đúng với mọi n.
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
Chứng minh:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{5.7.9}+\dfrac{36}{9.11.13}+...+\dfrac{36}{25.27.29}< 3\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\in< 1\left(n\in N,n\ge2\right)\)
\(D=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< 4\left(n\in N,n\ge2\right)\)
\(E=\dfrac{2!}{3!}+\dfrac{2!}{4!}+\dfrac{2!}{5!}+...+\dfrac{2!}{n!}< 1\left(n\in N,n\ge3\right)\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
Chứng minh rằng :
a) \(\dfrac{1.3.5.....39}{21.22.23.....40}=\dfrac{1}{2^{20}}\)
b) \(\dfrac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\dfrac{1}{2^n}\) với \(n\in\) N*
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
\(u_n=\dfrac{n+1}{2^{n+1}}\left(\dfrac{2}{1}+\dfrac{2^2}{2}+\dfrac{2^3}{3}+...+\dfrac{2^n}{n}\right)\).
Chứng minh \(\left(u_n\right)\) có giới hạn và tìm giới hạn đó.
Lời giải:
\(u_{n+1}=\frac{n+2}{2^{n+2}}\left(\frac{2}{1}+...+\frac{2^{n+1}}{n+1}\right)=\frac{n+2}{2^{n+1}}\left(\frac{2^{n+1}}{n+1}u_n+\frac{2^{n+1}}{n+1}\right)=\frac{n+2}{2n+2}(u_n+1)\)
Ta chứng minh $u_n\geq 1(*)$ với mọi $n=1,2,...$
Thật vậy:
$u_1=1; u_2=\frac{3}{2}>1$. Giả sử $(*)$ đúng đến $n=k$
$u_{k+1}=\frac{k+2}{2k+2}(u_k+1)>\frac{2(k+2)}{2k+2}>1$
Do đó $u_n\geq 1$ với mọi $n=1,2,...$
Tiếp theo ta chứng minh $u_n< 1+\frac{4}{n}(**)$ với mọi $n=1,2,...$
Thật vậy:
$u_1=1< 1+\frac{4}{1}$
$u_2=\frac{3}{2}< 1+\frac{4}{2};....;u_4=\frac{5}{3}<1+\frac{4}{4}$
....
Giả sử $(**)$ đúng đến $n=k\geq 5$. Khi đó:
\(u_{k+1}=\frac{k+2}{2k+2}(u_k+1)<\frac{k+2}{2k+2}(2+\frac{4}{k})=\frac{(k+2)^2}{k(k+1)}\)
\(\frac{(k+2)^2}{k(k+1)}-(1+\frac{4}{k+1})=\frac{(k+2)^2-k(k+5)}{k(k+1)}=\frac{4-k}{k(k+1)}<0\) với mọi $k\geq 5$
$\Rightarrow u_{k+1}< 1+\frac{4}{k+1}$. Phép quy nạp hoàn tất.
Do đó $(**)$ đúng
Từ $(*); (**)\Rightarrow 1\leq u_n\leq 1+\frac{4}{n}$ với mọi $n=1,2,...$
Mà $\lim (1+\frac{4}{n})=1$ khi $n\to +\infty$ nên $\lim u_n=1$
Chứng minh các mệnh đề sau:
\(a,1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\) \(\forall n\in N\) *
\(b,1.2+2.3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\) \(\forall n\in N\) *
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b:
Chứng minh với mọi số tự nhiên \(n\ge2\) :
\(M=\left(1-\dfrac{3}{2.4}\right).\left(1-\dfrac{3}{3.5}\right).\left(1-\dfrac{3}{4.6}\right).\left(1-\dfrac{3}{5.7}\right)...\left(1-\dfrac{3}{n\left(n+2\right)}\right)>\dfrac{1}{4}\)
\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)