a)5ⁿ⁺¹ -4.5n=5ⁿ.5-4.5ⁿ=5ⁿ(5-4)=5ⁿ

b)3xⁿ.(6x^n-3+1)-2xⁿ(9x^n-3-1)=18x^2n-3+3xⁿ-18^2n-3+2xⁿ

                                       =3xⁿ+2xⁿ=5xⁿ

D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

= \(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

= \(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)

= \(\frac{1}{5}.\frac{5n+5}{5n+6}\)

= \(\frac{n+1}{5n+6}\)

a: =1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100=49/100

b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)

=5/3*102/103

=510/309=170/103

c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)

=1/2*16/51=8/51

Chọn B.

a) \(\lim \frac{{5n + 1}}{{2n}} = \lim \frac{{5 + \frac{1}{n}}}{2} = \frac{{5 + 0}}{2} = \frac{5}{2}\)           

b) \(\lim \frac{{6{n^2} + 8n + 1}}{{5{n^2} + 3}} = \lim \frac{{6 + \frac{8}{n} + \frac{1}{{{n^2}}}}}{{5 + \frac{3}{{{n^2}}}}} = \frac{{6 + 0 + 0}}{{5 + 0}} = \frac{6}{5}\)                   

c) \(\lim \frac{{\sqrt {{n^2} + 5n + 3} }}{{6n + 2}} = \lim \frac{{\sqrt {1 + \frac{5}{n} + \frac{3}{{{n^2}}}} }}{{6 + \frac{2}{n}}} = \frac{{\sqrt {1 + 0 + 0} }}{{6 + 0}} = \frac{1}{6}\)

d) \(\lim \left( {2 - \frac{1}{{{3^n}}}} \right) = \lim 2 - \lim {\left( {\frac{1}{3}} \right)^n} = 2 - 0 = 0\)              

e) \(\lim \frac{{{3^n} + {2^n}}}{{{{4.3}^n}}} = \lim \frac{{1 + {{\left( {\frac{2}{3}} \right)}^n}}}{4} = \frac{{1 + 0}}{4} = \frac{1}{4}\)                       

g) \(\lim \frac{{2 + \frac{1}{n}}}{{{3^n}}}\)

Ta có \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} =  + \infty  \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)

Ta có:  u n = − 5 n 1 + 4 n − 5 n − 7 n − 7 + 4.4 n − 7 n = 5 7 n ⋅ 1 + − 4 5 n − 7 + 4. − 4 7 n

Vì lim − 4 5 n = lim − 4 7 n = 0  nên lim 1 + − 4 5 n − 7 + 4. − 4 7 n = − 1 7  và lim 5 7 n = 0

Do đó  lim u n = 0

Chọn đáp án D.