`Answer:`

 \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)

a) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)

\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)

b) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)

\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)

\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)

\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)

\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)

toán lớp 2

bt ko mà nói ^^

tui ghét lũy thừa nên lười

mik cx ko bt câu này

mik cx dg định đăng câu này

hok tốt

B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

Phải là \(\dfrac{6}{15}+\dfrac{6}{16}+...+\dfrac{6}{19}\) ms đúng

(1+3)+3²(1+3+3²+3³)+3⁶(1+3+3²+3³)+...+3⁹⁶(1+3+3²+3³)

=4+3².40+3⁶.40+....+3⁹⁶.40

=4+(3²+3⁶+....+3⁹⁶).40:40;4+(3²+3⁶+....3⁹⁶).40:4

S = 3¹⁰⁰ - 1

cs chép sai đè ko vậy

không

tự giải đi