rút gọn biểu thức
\(D=\left(n+1\right)\left(n^2+1\right)\left(n^{2^2}+1\right)\left(n^{2^3}+1\right)...\left(n^{2^m}+1\right)\)
Rút gọn biểu thức sau
\(D=n^2\left(n+4\right)\left(n-4\right)+\left(1-n^2\right)\left(n^2+1\right)\)
\(E=\left(\frac{1}{2}x^m-y^n\right)×\left(y^n+\frac{1}{2}x^m\right)\)
Giúp mình với !!!
rút gọn biểu thức :
N = 1 + \(\left(\dfrac{1}{2}\right)\) + \(\left(\dfrac{1}{2}\right)^2\) + \(\left(\dfrac{1}{2}\right)^3\) + ... + \(\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
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. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
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. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)
\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)
\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)
\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)
\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
Rút gọn biểu thức
\(A=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
Bài 1: \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)với n thuộc N; n lớn hơn hoặc bằng 2
Rút gọn thành biểu thức trên
A= \(\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)......\left(\frac{\left(n-1\right)\left(n+1\right)}{n.n}\right)\)
\(=\frac{3.8.15....\left(n-1\right)\left(n+1\right)}{\left(2.3.4......n\right)\left(2.3.4.......n\right)}=\frac{1.3.2.4.3.5.......\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4..................n\right)}=\frac{\left(1.2.3.......\left(n-1\right)\right)\left(3.4.5........\left(n+1\right)\right)}{\left(2.3.4.....n\right)\left(2.3.4...........n\right)}\)
\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)
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Rút gọn biểu thức sau:
\(A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+n}\right)\)với n lá các số tự nhiên lớn hơn 2.
Với mọi k thuộc N và k > 2 thì ta có :
\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)
Áp dụng vào A ta được :
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)
\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)
Rút gọn tổng: \(S=C\overset{1}{n}+1.2C\overset{2}{n}+2.3C\overset{3}{n}+...+\left(n-1\right)nC\overset{n}{n}\) bằng:
A. \(\left(n-1\right)n.2^{n-2}\)
B. \(n.2^{n-2}\)
C. \(\left(n-1\right)n.2^{n-1}+n\)
D. \(\left(n-1\right)n.2^{n-2}+n\)
Rút gọn tổng: \(S=C\overset{1}{n}+1.2C\overset{2}{n}+2.3C\overset{3}{n}+...+\left(n-1\right)nC\overset{n}{n}\) bằng:
A. \(\left(n-1\right)n.2^{n-2}\)
B. \(n.2^{n-2}\)
C. \(\left(n-1\right)n.2^{n-1}+n\)
D. \(\left(n-1\right)n.2^{n-2}+n\)
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Đạo hàm 2 vế:
\(n\left(1+x\right)^{n-1}=C_n^1+2xC_n^2+...+nx^{n-1}C_n^n\)
Tiếp tục đạo hàm 2 vế:
\(\left(n-1\right)n\left(1+x\right)^{n-2}=2C_n^2+2.3xC_n^3+...+\left(n-1\right)nx^{n-2}C_n^n\)
Thay \(x=1\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}=1.2C_n^2+2.3C_n^3+...+\left(n-1\right)nC_n^n\)
\(\Rightarrow\left(n-1\right)n.2^{n-2}+n=C_n^1+1.2C_n^2+...+\left(n-1\right)n.C_n^n\)
\(\Rightarrow S=\left(n-1\right)n.2^{n-2}+n\)