Chứng minh \(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{\left(4++\left(2n-1\right)\right)^4}=\frac{^{n^2}}{4n^2+1}\) 

1/(4+1^4)+3/(4+3^4)+...+(2n-1)/(4+(2n-1)^4)=n^2/(4n^2+1)

Tính N = 1^2/2^2-1 * 3^2/4^2-1 *** (2n+1)^2/(2n+2)-1

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

^{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)

cm bằng pp quy nạp P(n) = 2^2 +4^2+...+(2n) ^2=(2n(2n+1)(2n+1))/3

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

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

giup toi bai toan sau

Tinh: B= 1^2/2^2-1*3^2/4^2-1*5^2/6^2-1*...*(2n+1)^2/(2n+2)^2-1

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

o l m . v n

4,n^3-2 chia hết cho n-2

5, n^3-3n^2-3n-1 chia hết cho n^2+n+1

6, 5^n-2^n chia hết cho 63