Tính N = 1^2/2^2-1 * 3^2/4^2-1 *** (2n+1)^2/(2n+2)-1
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)
Tính các giới hạn sau
1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)
2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)
b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)
c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
tính tổng dãy tính sau
1)1+2+3+...+n
2)1+3+5+...+(2n+1)
3)2+4+6+...+2n
1) \(\frac{n\left(n+1\right)}{2}\)
2) \(\)
tính các giới hạn sau
a) lim (3n^2+n^2-1) b)lim n^3+3n+1/2n-n^3
c) lim -2n^3+3n+1/n-n^2 d) lim(n+ căn n^2-2n
e) lim (2n-3*2n+1) f) (căn 4n^2-n -2n) g) lim (căn n^2+3n-1 - 3^căn n^3-n)
Chụp ảnh hoặc sử dụng gõ công thức nhé bạn. Để vầy khó hiểu lắm
Tìm số nguyên n sao cho
a) (2n^3 + n^2 + 7n + 1) chia hết cho 2n-1
b)(n^3 - 2) chia hết cho n-2
c)(n^3 - 3n^2 - 3n -1) chia hết cho n^2 + n + 1
d)((n^4 - 2n^3 = 2n^2 - 2n + 1) chia hết cho n^4 - 1
e)(n^3 - n^2 + 2n + 7) chia hết cho n^2 + 1
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)
Tìm số nguyên n:
a) \(n^2+2n-4⋮11\)
b) \(2n^3+n^2+7n+1⋮2n-1\)
c) \(n^3-2⋮n-2\)
d) \(n^3-3n^2-3n-1⋮n^2+n+1\)
e) \(n^4-2n^3+2n^2-2n+1⋮n^4-1\)
g) \(n^3-n^2+2n+7⋮n^2+1\)
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
Giúp mình 2 bài này với
Bài 1: Tính Q= 1*3/3*5+2*4/5*7+3*5/7*9+...+(n-1)(n+1)/(2n-1)(2n+1)+....+1002*1004/2005*2007
Bài 2: tính R= 22/1*3+32/2*4+42/3*5+...+20062/2005*2007
tìm số nguyên n sao cho :
1,n^2+2n-4 chia hết cho 11
2,2n^3+n^2+7n+1 chia hết cho 2n -1
3,n^4-2n^3+2n^2-2n+1 chia hết cho n^4-1