\(2^2\cdot3^{2n}\cdot\left(\frac{2}{3}\right)^n\cdot2^n=82944\)
\(2^2\cdot\left(3^2\right)^n\cdot\left(\frac{2^n}{3^n}\right)\cdot2^n=82944\)
\(2^2\cdot9^n\cdot\frac{2^n}{3^n}\cdot2^n=82944\)
\(2^2\cdot\frac{9^n\cdot2^n}{3^n}\cdot2^n=82944\)
\(2^2\cdot\frac{18^n}{3^n}\cdot2^n=82944\)
\(4\cdot6^n\cdot2^n=82944\)
\(6^n\cdot2^n=82944:4\)
\(12^n=20736\)
\(12^n=12^4\)
Vậy n=4
cho A=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+........+\dfrac{2n-1}{\left(n-1\right)^2.n^2}\)
CMR:A<1
Chứng minh rằng với mọi n thuộc Z thì :
a) \(\left(n^2+3n-1\right).\left(n+2\right)-n^3+2⋮5\)
b) \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
c) \(\left(2n-1\right).3-\left(2n-1\right)⋮8\)
d) \(n^2\left(n+1\right)+2n\left(n+1\right)⋮6\)
Tính tổng: \(S=\dfrac{3}{1^2.3}+\dfrac{5}{\left(1^2+2^2\right).4}+\dfrac{7}{\left(1^2+2^2+3^2\right).5}+...+\dfrac{2n+1}{\left(1^2+2^2+3^2+...+n^2\right).\left(n+2\right)}\)
(mấy chế làm giùm đi , nhanh nhanh giùm nhen,mai học rồi)
CMR với mọi số tự nhiên \(n\ge1\):
a ) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2}\)
b ) \(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{4}.\)
Tính :
a) \(\frac{\left(\frac{-5}{7}\right)^n}{\left(\frac{-5}{7}\right)^{n-1}}\)( n\(\ge\)1 )
b) \(\frac{\frac{-1}{2}^{2n}}{\left(\frac{-1}{2}\right)^n}\) ( n \(\in\)N )
Chứng minh vs mọi n thuộc Z thì:
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)
\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
CMR:\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^2}< 1\)
Với mọi số tự nhiên n \(\ge\) 2 . Hãy so sánh
A= \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}với\frac{1}{2}\)
Tìm \(n\in Z\) sao cho:
\(a.\left(3n+1\right)⋮\left(2n+3\right)\)
\(b.\left(n^2+5\right)⋮\left(n+1\right)\)