C/M: 1+2+2^2+2^3+...+2^99+2^100
giải giúp mình nhé :)))
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29 tháng 6 2023
3A=1+1/3+...+1/3^99
=>2A=1-1/3^100=(3^100-1)/3^100
=>A=(3^100-1)/(2*3^100)
TT
0
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 5 2016
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
DM
2
20 tháng 4 2020
- Nhận xét:
Với mọi số dương n, ta có:
1 - n2 = ( 1 - n ) + ( n - n2 )
= ( 1 - n ) + n ( 1 - n )
= ( 1 - n )( 1 + n ).
Do đó:
\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\) \(\left(\frac{1}{99^2}-1\right)\)
= \(\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}.....\frac{1-99^2}{99^2}\)
= \(\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.\)\(\frac{\left(1-4\right)\left(1+4\right)}{4^2}.....\frac{\left(1-99\right)\left(1+99\right)}{99^2}\)
= \(\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}.....\frac{\left(-98\right).100}{99^2}\)
= \(\frac{\left(-1\right).3.\left(-2\right).4.\left(-3\right).5.....\left(-98\right).100}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(99.99\right)}\)
= \(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right).....\left(-98\right)\right]\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)
= \(\frac{\left(1.2.3.....98\right).\left(3.4.5.....100\right)}{\left(2.3.4.....99\right)\left(2.3.4.....99\right)}\)
= \(\frac{1.100}{99.2}\)
= \(\frac{50}{99}\).
20 tháng 4 2020
\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{99^2}-1\right)\)
\(=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{9801}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9800}{9801}=\frac{-\left(1.3\right)}{2.2}.\frac{-\left(2.4\right)}{3.3}.\frac{-\left(3.5\right)}{4.4}...\frac{-\left(98.100\right)}{99.99}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}=\frac{1.3.2.4.3.5...97.99.98.100}{2.2.3.3.4.4...99.99}\)
\(=\frac{\left(1.2.3.4....98\right).\left(3.4.5.6...100\right)}{\left(2.3.4.5...99\right).\left(2.3.4.5...99\right)}=\frac{100}{99.2}=\frac{50}{99}\)
LT
1
BT
0
1 tháng 1 2018
Ta có:
B=1/2-1/2^2-1/2^3-...-1/2^100
B/2=1/2^2-1/2^3-1/2^4-....-1/2^101
B/2-B=1/2^101-1/2
=>B=(1/2^101-1/2).2
Vậy:B=(1/2^101-1/2).2
6 tháng 9 2018
Làm trc cho 2 câu cuối
c) \(a^2-b^2-4a+4b\)
\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)
\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)
d) \(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)+1\)
\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)
thiếu phần trên là: " chia hết cho 3 "