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13 tháng 9 2015

c;=(50-49)(50+49)+(48-47)(48+47)+.............+(2+1)(2-1)

=50+49+48+............+1

=(50+1)50=2550:2=1275

d;=(2^4-1)(2^4+1)(2^8+1)(2^16+1)

=(2^8-1)(2^8+1)(2^16+1)

=(2^16-1)(2^16+1)

=2^32-1

e;=(3-1)(3+1)(3^2+1)...........(3^16+1)

=(3^2-1)(3^2+1)..............(3^16+1)

=(3^16-1)(3^16+1)=3^32-1

tu tinh ket qua luy thua tao khong thua hoi dau

10 tháng 7 2018

Đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)

\(A=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{10.11}{2}}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(A=2\cdot\frac{9}{22}=\frac{9}{11}\)

Vậy A = \(\frac{9}{11}\)

10 tháng 7 2018

Trả lời:

\(\frac{9}{11}\)nha

Chúc bạn học tốt

11 tháng 7 2023

a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)

\(=1-\dfrac{1}{256}\)

\(=\dfrac{255}{256}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{13}{14}\)

c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)

\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)

\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)

\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=\dfrac{1}{15}-\dfrac{1}{90}\)

\(=\dfrac{6}{90}-\dfrac{1}{90}\)

\(=\dfrac{5}{90}=\dfrac{1}{18}\)

 

11 tháng 7 2023

tớ đang cần gấp

 

26 tháng 10 2023

$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$

$=\dfrac{3^{32}-1}{2}$

11 tháng 5 2019

S=1+1/1+2+1/1+2+3+4+.....+1/1+2+3+4+5+6+7+8

S=1+1/1+2+1/1+2+3+4+1/1+2+3+4+5+6+1/1+2+3+4+5+6+7+8

S=1+3+10+21+36

S=71