CMR: \(13^n-1⋮12\left(\forall n\inℕ\right)\)

a, chứng minh:

\(n^4+\frac{1}{4}=\left[\left(n-1\right)n+\frac{1}{2}\right].\left[\left(n+1\right)n+\frac{1}{2}\right]\)

b, Áp dụng câu a) thu gọn:

\(\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right)...\left(13^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right)...\left(14^4+\frac{1}{4}\right)}\)

^{Cho \(n^4+\frac{1}{4}=\left(\left(n-1\right)n+\frac{1}{2}\right)\left(\left(n+1\right)n+\frac{1}{2}\right)\)}

^{Thu gọn phân thức:}

^{\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(13^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(14^4+\frac{1}{4}\right)}\)}

1. Chứng minh rằng với n là stn khác 0 thì \(4^{2n+1}+3^{n+2}\)chia hết cho 13.

2.Tính: 

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

CTR: \(\frac{1}{5}+\frac{1}{13}+\frac{1}{15}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}\)với mọi \(n\in N\)

Chứng minh: \(1\cdot2\cdot3+2\cdot3\cdot4+...+n\left(n+1\right)\left(n+2\right)=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)  với mọi \(n\inℕ\)

Tìm các số nguyên n thỏa mãn :

a)\(\left(n+5\right)⋮\left(n-2\right)\)

b)\(\left(2n+1\right)⋮\left(n-5\right)\)

c) \(\left(n^2+3n-13\right)⋮n+3\)

d)\(\left(n^2+3\right)⋮\left(n-1\right)\)

Tính:
a, \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{n^2\left(n+1\right)^1}\) tại n= 2014
b, \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{12}{13!}\)

1/ Tính

a, \(S=x^2-x^3+x^4-x^5+...+\left(-1\right)^nx^n....\)Với \(\left|x\right|< 1,n\ge2\)

b, Giai phương trình: \(2x+1+x^2-x^3+x^4-x^5+...+\left(-1\right)^nx^n....=\frac{13}{6}\)