Cho n + 1 tia chung gốc. Hỏi tạo được bao nhiêu góc?
\(\dfrac{\left(n+1\right)n}{2}\)
\(\dfrac{\left(n-1\right)n}{2}\)
(n +1 ).n
\(\dfrac{n^2}{2}\)
1 . Rút gọn : \(\left(\dfrac{2006}{1}+\dfrac{2015}{2}+....+\dfrac{2}{2015}+\dfrac{1}{2016}\right)\): \(\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
2 . a , Vẽ n tia chung gốc ( n \(\in\)N* ), chúng tạo thành 28 góc . Hỏi giá trị của n bằng bao nhiêu ?
b , Cho 1 số tia chung gốc tạo thành một số góc . Sau khi vẽ thêm tia chung gốc thì số góc tăng thêm là 9 . Hỏi lúc đàu có bao nhiêu tia ?
ai làm nhanh và đúng mk tick cho
1.a) A= \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{n-1}-1\right).\left(\dfrac{1}{n}-1\right),n\)thuộc N*
b) B= (\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{n^2}-1\right)\); n thuộc N*
Lời giải:
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)
\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)
\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)
b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)
\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)
\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)
\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)
\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)
Giải típ hộ mik nha !
\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2016}{2017}=\dfrac{n+1-2}{2.\left(n+1\right)}=\dfrac{2016}{2017}=\dfrac{n-1}{2.\left(n+1\right)}=\dfrac{2016}{2017}=2017.\left(n-1\right)=2016.2\left(n+1\right)=...\)
Tính các giới hạn
a) \(lim\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}\)
\(lim\left(n^3+n\sqrt{n}-5\right)\)
Giúp mình với ạ
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/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)
Tính giới hạn sau:
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^3}\left(1+2^2+...+\left(n-1\right)^2\right)\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}[\left(x+\dfrac{a}{n}\right)+\left(x+\dfrac{2a}{n}\right)+...+\left(x+\dfrac{\left(n-1\right)a}{n}\right)]\)
3) \(\lim\limits_{n\rightarrow\infty}\dfrac{1^3+2^3+...+n^3}{n^4}\)
1.
Trước hết bạn nhớ công thức:
$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)
Áp vào bài:
\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)
\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)
\(=1.\frac{1}{3}=\frac{1}{3}\)
2.
\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)
\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)
\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)
\(=x+a\)
3.
Trước tiên ta có công thức:
$1^3+2^3+....+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}$
Chứng minh: https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/
Khi đó:
\(\lim \frac{1^3+2^3+...+n^3}{n^4}=\lim \frac{n^2(n+1)^2}{4n^4}\\ =\lim \frac{(n+1)^2}{4n^2}=\frac{1}{4}\lim (1+\frac{1}{n})^2=\frac{1}{4}.1=\frac{1}{4}\)
Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:
A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)
C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:
A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)
C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)
\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)
\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)
\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)
\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
Cho dãy xác định \(\left\{{}\begin{matrix}u\left(1\right)=\dfrac{1}{4}\\u\left(n+1\right)=\left(u\left(n\right)\right)^2+\dfrac{u\left(n\right)}{2}\end{matrix}\right.\)
CM với mọi n thì 0<u(n)<\(\dfrac{1}{4}\) và\(\dfrac{u\left(n+1\right)}{u\left(n\right)}\le\dfrac{3}{4}\)
Từ đó suy ra limu(n)=o
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.