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TUI SẼ KO BAO GIỜ TICK CHO Đào Chiến Thắng

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a/ \(\lim\limits\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}=\lim\limits\dfrac{\dfrac{\left(\dfrac{1}{3}\right)^{n+1}-1}{\dfrac{1}{3}-1}}{\dfrac{\left(\dfrac{1}{2}\right)^{n+1}-1}{\dfrac{1}{2}-1}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3\)

b/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)

1, Vì n+2016, n+2017,n+2018 là 3 số tự nhiên liên tiếp nên tích chia hết cho 3

2, n²\(⋮\)n+1 (1)

Vì n+1\(⋮\)n+1   => (n+1)(n-1)\(⋮\)n+1

=> n²-1\(⋮\)n+1 (2)

Lấy (1) trừ (2) ta có   1\(⋮\)n+1

=>n+1=1=> n=0

\(\left(\frac{x-1}{x+2}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)^2+3\left(\frac{x+1}{x-2}\right)^2=0\left(1\right)\)

\(ĐKXĐ:x\ne\pm2\)

Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)

=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)

\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)

=>  \(b=0;a=0\)

Bạn cùng trường :">

\(1^2+2^2+3^2...+n^2=1+2\left(1+1\right)+3\left(2+1\right)+...+n\left(n-1+1\right)\\ =1+1\cdot2+2+3\cdot2+3+...+n\left(n-1\right)+n\\ =\left(1+2+3+...+n\right)+\left[1\cdot2+2\cdot3+...+n\left(n-1\right)\right]\)

Ta có \(1\cdot2+2\cdot3+...+n\left(n-1\right)\)

\(=\dfrac{1}{3}\left[1\cdot2\cdot3+2\cdot3\cdot3+...+3n\left(n-1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+n\left(n-1\right)\left(n+2+n+1\right)\right]\\ =\dfrac{1}{3}\left(1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)\right)\\ =\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\)

\(\Rightarrow1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\\ =\dfrac{3n\left(n+1\right)+2n\left(n-1\right)\left(n+1\right)}{6}=\dfrac{n\left(n+1\right)\left(3+2n-2\right)}{6}\\ =\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)

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.

\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)

\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)

\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)

\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)