Chọn câu trả lời đúng : 1.2 + 2.3 + 3.4 + ... + (n - 1)n + n(n + 1) =
A n(n+1)(2n+1)4+n(n+1)2
B n(n+1)(2n+1)6+n(n−1)2
C n(n+1)(2n+1)6+n(n+1)2
D n(n+2)(2n+1)6+n(n+1)2
chứng minh rằng
1, 1/n(n+1)=1/n-1/n+1
2, 2/n(n+1)(n+2)=1/n(n+1)-1/(n+1)(n+2)
3, 3/n(n+1)(n+2)(n+3)=1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3)
4, 4/(2n-1)(2n+1)(2n+3)=1/(2n+1)(2n-1)-1/(2n+1)(2n+3)
5, m/n(n+m)=1/n-1/n+m
6, 2m/n(n+m)(n+2n)=1/n(n+m)-1/(n+m)(n+2n)
Tìm giới hạn các dãy số sau
a) \(lim\dfrac{2^n+6^n-4^{n-1}}{3^n+6^{n+1}}\)
b) \(lim\dfrac{1+3+5+...+\left(2n+1\right)}{3n^2+4}\)
c) \(lim\dfrac{1+2+3+...+n}{n^2-3}\)
d) \(lim\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right]\)
e) \(lim\left[\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
a, tính tổng:1+2+3+...+n ,1+3+5+...+(2n-1)
b, tính tổng: 1.2+2.3+3.4+...+n.(n+1)
1.2.3+2.3.4+3.4.5+...+n(n+1).(n+2)
Chứng minh các mệnh đề sau:
\(a,1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\) \(\forall n\in N\) *
\(b,1.2+2.3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\) \(\forall n\in N\) *
1.Chứng minh các dạng tổng quát sau:
a) An=-1+3-5+7-9+11-...+(-1^n)(2n-1)
b)Dn=1/1.2+1/2.3+1/3.4+...+1/n(n+1)=n(n+1)(n+2)(n+3)/4
giúp mình nhé mình sắp phải nộp rồi.Ai trả lời đúng mình sẽ k cho
chứng minh rằng
1, 1/n(n+1)=1/n-1/n+1
2, 2/n(n+1)(n+2)=1/n(n+1)-1/(n+1)(n+2)
3, 3/n(n+1)(n+2)(n+3)=1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3)
4, 4/(2n-1)(2n+1)(2n+3)=1/(2n+1)(2n-1)-1/(2n+1)(2n+3)
5, m/n(n+m)=1/n-1/n+m
6, 2m/n(n+m)(n+2n)=1/n(n+m)-1/(n+m)(n+2n)
ai nhanh mình tick trước 9 giờ
\(1) VP= \frac{1}{n}-\frac{1}{n+1}\)\(= \frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\)\(= \frac{n+1-n}{n(n+1)}\)\(= \frac{1}{n(n+1)}\)\(= VT\)
2) \(VP= \frac{1}{n+1}-\frac{1}{(n+1)(n+2)}= \frac{(n+2)}{n(n+1)(n+2)}-\frac{n}{n(n+1)(n+2)}\)\(= \frac{n+2-n}{n(n+1)(n+2)}= \frac{2}{n(n+1)(n+2)}=VT\)
3) \(VP= \frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}=\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\)\(= \frac{n+3-n}{n(n+1)(n+2)(n+3)}=\frac{3}{n(n+1)(n+2)(n+3)(n+4)}=VT\)
Những ý sau làm tương tự, thế mà chẳng thèm mở mồm ra hỏi bạn :))
chị thương ơi gửi em đề bài câu 9,10 toán bài 2
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(\Rightarrow F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(\Rightarrow F=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\left(đpcm\right)\)
\(H=2+4+6+...+2n\)
Tính giá trị các tổng sau theo n:(n>0)
A=1+2+3+....+n
B=1+3+5+...+(2n+1)
C=1.2+2.3+3.4+.....+n(n+1)
D=1.2.3+2.3.4+3.4.5+....+n(n+1)(n+2)
1. CMR: ∀ n∈\(N^{\cdot}\)
a) \(A=5^n+2.3^{n-1}+1\text{⋮}8\)
b) \(B=3^{n+2}+4^{2n+1}\text{⋮}13\)
c) \(C=6^{2n}+3^{n+2}+3^n\text{⋮}11\)
d) \(D=1^n+2^n+5^n+8^n\text{⋮}8\)
2. \(CMR:\) \(1^{2002}+2^{2002}+...+2002^{2002}\text{⋮}11\)
3. a) cho a,b ∈Z, t/m:\(a^2+b^2\text{⋮}7\). \(CMR:a\text{⋮}7;b\text{⋮}7\)
b) \(CMR:\) Nếu \(a^2+b^2\text{⋮}21\) thì \(a^2+b^2\text{⋮}441\) (a,b ∈Z)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
\(2,\)
Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)
Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)