cho\(n\inℕ^∗\)
CMR:\(\left(n^4+2015\cdot n^2\right)⋮12\)
Cho đa thức: f( x ) = \(2\cdot\left(x^2\right)^n-5\cdot\left(x^n\right)^2+8\cdot x^{n-1}\cdot x^{1+n}-4\cdot x^{n^2+1}\cdot x^{2\cdot n-n^2-1}\left(n\inℕ\right)\)
a, Thu gọn đa thức f(x)
b, Tìm giá trị nhỏ nhất của f(x) + 2020
a) \(f\left(x\right)=2.\left(x^2\right)^n-5.\left(x^n\right)^2+8n^{n-1}.x^{1+n}-4.x^{n^2+1}.x^{2n-n^2-1}\)
\(=2x^{2n}-5x^{2n}+8x^{2x}-4x^{2n}\)
\(=x^{2n}\)
b) \(f\left(x\right)+2020=x^{2n}+2020\)
Vì \(n\in N\Rightarrow2n\in N\)và 2n là số chẵn
\(\Rightarrow x^{2n}\ge1\)
\(\Rightarrow x^{2n}+2020\ge2021\)
Dấu"="xảy ra \(\Leftrightarrow x^{2n}=1\)
\(\Leftrightarrow n=0\)
Vậy ...
( ko bít đúng ko -.- )
Chứng minh rằng:
a)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)
b)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot\cdot\cdot2n}=\frac{1}{2^n}\)Với \(n\inℕ^∗\)
CMR: \(13^n-1⋮12\left(\forall n\inℕ\right)\)
\(Ta có : 13^n - 1\)
\(= ( 13 - 1 )( 13\)\(n - 1\) \(+ 13\)\(n - 2\) \(+ ... + 13 . 1\)\(n - 2\) \(+1\)\(n - 1\) \()\)
\(= 12 . ( 13\)\(n - 1\) \(+ 13\)\(n - 2\)\(.1 + ... + 13 . 1\)\(n - 2\) \(+ 1\)\(n - 1\)\()\)\(⋮\)\(12\)
\(Vậy : 13^n - 1 \)\(⋮\)\(12\)
Chứng minh :
\(\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot....\cdot2n\) \(⋮\) \(2^n\) \(\left(n\inℕ^∗\right)\)
Với n = 1 => Ta có: (1+1) = 2 chia hết cho 21
Giả sử n = k thì (k+1).(k+2)...2k chia hết cho 2k
Cần chứng minh: (k+1+1).(k+1+2)...2(k+1) chia hết cho 2k+1
Ta có: (k+1+1).(k+1+2)...2(k+1) = (k+2).(k+3)....2k.2(k+1) = 2.(k+1).(k+2)...2k chia hết cho 2.2k = 2k+1
Vậy (n+1)(n+2)....2n chia hết cho 2n (với mọi n thuộc N*)
Nhân \(\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n\) với \(2.4.6.8...2n\)
Ta được: \(\left(2.4.6...2n\right)\left(n+1\right)\left(n+2\right)...2n\)
=\(\left(1.2.3..n\right).2^n\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n⋮2^n\)
\(\Rightarrow\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n⋮2^n\)
\(f\left(n\right)=\left(n^2+n+1\right)^2+1\). Xét dãy \(\left(u_n\right)\) sao cho : \(\left(u_n\right)=\dfrac{f\left(1\right)\cdot f\left(3\right)\cdot f\left(5\right)...\cdot f\left(2n-1\right)}{f\left(2\right)\cdot f\left(4\right)\cdot...\cdot f\left(2n\right)}\). Tính \(\lim\limits_{n\sqrt{u_n}}\)
CMR \(n\inℕ^∗\)thì \(\left(5^{2n+1}+2^{n+4}+2^{n+1}\right)⋮23\)
Cho \(n\inℕ^∗\)CMR
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=1+\frac{1}{n}-\frac{1}{\left(n+1\right)}\)
\(\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2-2\left(\frac{1}{n}-\frac{1}{n\left(n+1\right)}-\frac{1}{n+1}\right)}\)
=1+1/n-1/n+1
chúc bn hoc tốt
Cho \(n\inℕ^∗\) CMR
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=1+\frac{1}{n}-\frac{1}{\left(n+1\right)}\)
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=1+\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=\frac{[\left(n+1\right)^2-n]^2}{n^2\left(n+1\right)^2}\)
\(\Rightarrow\left(n+1\right)^4+n^2=\left(n+1\right)^4-2\left(n+1\right)^2n+n^2\)
\(\Rightarrow0=-2\left(n+1\right)^2n\)
\(\Rightarrow\orbr{\begin{cases}\left(n+1\right)^2=0\\n=0\end{cases}}\Rightarrow\orbr{\begin{cases}n=-1\\n=0\end{cases}}\) mà \(n\inℕ^∗\)
=> n\(\in\varnothing\)
Ui nhầm ! sr bạn nha , tội ẩu ko đọc kĩ đề :(
cho f(n)=(n2 + n +1 )2 +1 với n thuộc N* . Đặt \(p_n=\frac{f_{\left(1\right)}\cdot f_{\left(3\right)}\cdot f_{\left(5\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n-1\right)}}{f_{\left(2\right)}\cdot f_{\left(4\right)}\cdot f_{\left(6\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n\right)}}\)
chứng minh rằng : P1 + P2 +P3 +................+ Pn <1/2