Let \(S_n=\Sigma^n_{k=1}k!\left(k^2+3k+1\right)\left(n\inℕ^∗\right)\)
Prove that \(S_{400}\equiv2002\left(mod2005\right)\)
Let \(a,b,c,k\) be positive real numbers such that \(k\left(ab+bc+ca\right)+2abc\le k^3\) . Prove that:
\(\left(1\right)k\left(a+b+c\right)\ge2\left(ab+bc+ca\right)\)
\(\left(2\right)k\left(a^3+b^3+c^3\right)\ge2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\left(3\right)k\left(a^{2n-1}+b^{2n-1}+c^{2n-1}\right)\ge2\left(a^nb^n+b^nc^n+c^na^n\right)\) \(\left(n\ge0;n\in R\right)\)
mày bị điên rồi hả câu hỏi thế này làm gì có người giải được
CMR:nếu \(1+2^n+4^n\) là số nguyên tố \(\left(n\inℕ^∗\right)\) thì n=3k \(\left(k\inℕ^∗\right)\)
Chứng minh rằng với k \(\in\) N* ta luôn có \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\)
Ta có:
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\\ =k\left(k+1\right)\left[\left(k-2\right)-\left(k-1\right)\right]\\ =k\left(k+1\right)\left[k-2-k+1\right]\\ =k\left(k+1\right)\left\{\left[k+\left(-k\right)\right]+\left(2+1\right)\right\}\\ =k\left(k+1\right).3\\ =3.k\left(k+1\right)\)
Vậy \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\\ =3.k.\left(k+1\right)\)
Ta có:
\(VT=k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\)
\(=k\left(k+1\right)\left[\left(k+2\right)-\left(k-1\right)\right]\)
\(=k\left(k+1\right)\left[k+2-k+1\right]\)
\(=k\left(k+1\right)\left[\left(k-k\right)+\left(2+1\right)\right]\)
\(=k\left(k+1\right).3\)
\(=3k\left(k+1\right)\)
\(\Rightarrow VT=VP\)
Vậy với \(k\in N\)* thì ta luôn có:
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\) (Đpcm)
Chứng tỏ: \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\)
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\)
\(VT=\left(k+1\right)\left[k\left(k+2\right)-k\left(k-1\right)\right]=\left(k+1\right)\left(k^2+2k-k^2+k\right)\)
\(=\left(k+1\right).3k=VP\)
Tính \(lim\dfrac{\prod\limits^n_{k=1}\left(2k-1\right)}{\prod\limits^n_{k=1}\left(2k\right)}\)
Bạn tham khảo cách làm nha
https://diendantoanhoc.org/topic/106253-lim-nto-inftyprod-k1nfrac2k-12k/
chứng minh: \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\)
trong đó k thuộc N*
từ đó suy ra công thức tính tổng
\(S=1.2+2.3+3.4+...+n\left(n+1\right)\)
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=k\left(k+1\right)\left[\left(k+2\right)-\left(k-1\right)\right]=3k\left(k+1\right)\)
Công thức tinh tổng là : \(S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=k\left(k+1\right)\left(k+2-k+1\right)=3k\left(k+1\right)\left(ĐPCM\right)\)
\(S=1.2+2.3+3.4+...+n\left(n+1\right)\)
3\(S=3\left[1.2+2.3+3.4+...+n\left(n+1\right)\right]\)
\(3S=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
3S=n(n+1)(n+2)
\(S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Cho \(S=1.2.3+2.3.4+3.4.5+...+k\left(k+1\right)\left(k+2\right)\) với \(\left(k\inℕ^∗\right)\). Chứng minh rằng \(4S+1\) là bình phương của một số tự nhiên.
4S=1.2.3.4+2.3.4.4+3.4.5.4+...+k(k+1)(k+2).4=
=1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+k(k+1)(k+2)[(k+3)-(k-1)]=
=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-(k-1)k(k+1)(k+2)+k(k+1)(k+2)(k+3)=
=k(k+1)(k+2)(k+3)=k(k+3)(k+1)(k+2)=
=(k2+3k)(k2+3k+2)=(k2+3k)2+2(k2+3k)
=> 4S+1=(k2+3k)2+2(k2+3k)+1=[(k2+3k)+1]2
\(S_k=\left(\sqrt{2}+1\right)^k+\left(\sqrt{2}-1\right)^k\), k thuộc N
Chứng minh \(S_{2019}.S_{2010}-S_{4019}=2\sqrt{2}\)
Đề là \(S_{2009}.S_{2010}\) chứ
Đặt \(\sqrt{2}+1=a;\sqrt{2}-1=b\Rightarrow ab=1\)
Ta có: \(S_{2009}.S_{2010}=\left(a^{2009}+b^{2009}\right)\left(a^{2010}+b^{2010}\right)\)
\(=a^{2009}.a^{2010}+b^{2009}.a^{2010}+a^{2009}.b^{2010}+b^{2009}.b^{2010}\)
\(=a^{2009}.b^{2009}\left(a+b\right)+a^{4019}+b^{4019}\)
\(=1.2\sqrt{2}+S_{4019}=S_{4019}+2\sqrt{2}\)
\(\Rightarrow S_{2009}.S_{2010}-S_{4019}=2\sqrt{2}\)
Rút gọn :
a, \(A=\sum\limits^n_{k=1}k.k!\)
b, \(B=\sum\limits^n_{k=2}\dfrac{k}{\left(k-1\right)!}\)