1/ (6⁹.2¹⁰+12¹⁰)+(2¹⁹.27³+15.4⁹.9⁴)  = 2⁹.3⁹.2¹⁰+3¹⁰.2²⁰+2¹⁹.3⁹+5.3.2¹⁸.3⁸ = 2¹⁹.3⁹+3¹⁰.2²⁰+2¹⁹.3⁹+5.2¹⁸.3⁹

⁼ 2¹⁸.3⁹(2+3.2²+5)=19.2¹⁸.3⁹

sao bạn lại nhắn vớ va vớ vậy PHẠM ĐỨC PHÚC

1/ (69
.210+1210
)+(219
.273+15.49
.94
)  = 29
.39
.210+310
.220+219
.39+5.3.218
.38
 = 219
.39+310
.220+219
.39+5.218
.39
= 2
18
.39
(2+3.22+5)=19.218
.39

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{n\left(n+1\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{n\left(n+1\right)}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{n\left(n+1\right)}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{n\left(n+1\right)}\right)\)

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

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

\(=1-\frac{2}{n+1}\)

\(=\frac{n+1}{n+1}-\frac{2}{n+1}\)

\(=\frac{n-1}{n+1}\)

bài 2 x đâu vậy bn

viết cả cách làm nhé!

Bài 1:

a. https://olm.vn/hoi-dap/detail/100987610050.html

b. Giống nhau hoàn toàn => P=Q

Chỉ biết thế thôi

a,1−5−9+13+17−21−25+29+......+2001−2005−2009+2013=(1−5)−(9−13)+(17−21)−(25−29)+.....+(2001−2005)−(2009−2013)=−4+4−4+4−......−4+4=0mik biết làm z thoy

\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)

\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-1+\frac{1}{100}\)

\(A=\frac{2}{100}-1\)

\(A=\frac{1}{50}-1\)

\(A=\frac{-49}{50}\)

b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\)        (1)

Đặt \(B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\)

\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)

             \(=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}\)

\(2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)\)

                 \(=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)\)

             \(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)

                 \(=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3\)

Đặt \(C=2^3+2^4+2^5+2^n\)

\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)

         \(C=2^4+2^5+2^6+...+2^{n+1}\)

\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)

\(C=2^{n+1}-2^3\)

Khi đó :  \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)

                  \(=-2^{n+1}+2^3+n.2^{n+1}-2^3\)

                   =\(=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}\)

Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)

                           \(2^{n+34}-\left(n-1\right).2^{n+1}=0\)

                          \(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)

Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))

\(n=2^{33}+1\)

Vậy \(n=2^{33}+1\)

a=\(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{100}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)

=>b/a=2011

hình như đề : CMR : \(\frac{b}{a}\)là 1 số nguyên

Ta có :

\(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(a=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(a=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(a=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(a=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(a=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(b=\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}\)

\(b=2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)

\(\Rightarrow\frac{b}{a}=\frac{2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=2011\)là 1 số nguyên ( đpcm )

Sửa đề: Chứng minh \(\frac{b}{a}\)là một số nguyên

Ta có: \(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

Áp dụng quy tắc dấu ngoặc vào tổng đại số trên , và theo quy luật của tổng đại số.ta có:

\(a=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

Tiếp tục phân tích , ta được:

\(a=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Ta có: \(\frac{b}{a}=\frac{\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=\frac{2011\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=\frac{2011}{1}=2011\)là một số nguyên (đpcm)

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{2}{n\left(n+1\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2010}{2011}\)

\(\Leftrightarrow n=4021\).