Rút gọn biểu thức
a, \(3^{n+2}-3^{n+1}+6.3^n\)
b, \(\left(3.2^{n+2}+2^n+2^{n+1}\right):5\)
rút gọn : a) 3^n+2-3^n-1-6.3^n. b) (3.2^n+3+2^n-2^n-1):5
Rút gọn:
a) 3^n+2 - 3^n+1 - 6.3^n
b) (3.2^n+2+2^n -2^n-1) : 5
a)3^n+2-3^n+1-6.3^n=3^n.9-3^n.3+6.3^n
=3^n(9-4+6)
=3^n.11
1.Rút gọn biểu thức:
a)3n+2-3n+2+6.3n
b)(3.2n-2+2n-2n+2):5
2.Chứng minh rằng với mọi số nguyên n thì (3n+2-2n+2+3n-2n-) chia hết cho 10
a)\(3^{n+2}-3^{n+2}+6.3^n=0+6.3^n\)
b)
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 \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) 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
b- \(\sqrt{1+2+3+....+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(A=1+2+...+\left(n-1\right)=\frac{n\left(n-1\right)}{2}\)
\(B=\left(n-1\right)+..+2+1=\frac{\left(n-1\right)n}{2}\)
\(A+n+B=\frac{\left(n-1\right)n}{2}+n+\frac{\left(n-1\right)n}{2}=\left(n-1\right)n+n=n^2\)
n là tự nhiên \(\sqrt{n^2}=n\)
Rút gọn biểu thức
\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
với n thuộc N
Ta có : \(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}=\sqrt{2\left(1+2+3+...+n-1\right)+n}\)
\(=\sqrt{2\left(n-1\right).\left(n-1+1\right):2+n}=\sqrt{\left(n-1\right).n+n}=\sqrt{\left(n-1+1\right).n}=\sqrt{n^2}=n\)
rút gọn biểu thức
A=\(\dfrac{Pn+2}{A^k_n.Pn-k}+\dfrac{C^5_{15}+2.C^6_{15}+C^7_{15}}{C^7_{17}}\)
B=\(\dfrac{Pn-Pn-1}{\left(n-2\right)!}\)
\(A=\dfrac{n!+2}{\dfrac{n!}{\left(n-k\right)!}\cdot n!-k}+\dfrac{3003+10010+6435}{19448}\)
\(=\dfrac{n!+2}{n\left(n-1\right)\cdot...\cdot\left(n-k+1\right)\cdot n!-k}+1=\dfrac{n!+2+\dfrac{n!^2}{\left(n-k\right)!}-k}{\dfrac{n!^2}{\left(n-k\right)!}-k}\)
\(B=\dfrac{n!-\left(n-1\right)!}{\left(n-2\right)!}=\dfrac{\left(n-1\right)!\left(n-1\right)}{\left(n-2\right)!}=\left(n-1\right)^2=n^2-2n+1\)