Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
E=12+22+32+42+...+982+992+1002
=1+2(1+1)+3(1+2)+4(1+3)+....+98(1+97)+99(1+98)+100(1+99)
=1+1.2+2+3+2.3+4+3.4+....+98+97.98+99+98.99+100+99.100
=(1+2+3+4+...+100)+(1.2+2.3+3.4+...+99.100)
Đặt A=1+2+3+...+100=\(\frac{\left(100+1\right).100}{2}=5050\)
Đặt B=1.2+2.3+3.4+...+99.100
3B=1.2.3+2.3.3+....+99.100.3
3B=1.2.3+2.3.(4-1)+...+99.100.(101-98)
3B=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100
3B=99.100.101
=>B=\(\frac{99.100.101}{3}=333300\)
Vậy E=A+B=5050+333300=338350
\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.......+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(\Leftrightarrow2B=1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+........+\dfrac{98}{2^{99}}+\dfrac{99}{2^{100}}\)
\(\Leftrightarrow2B-B=\left(1+\dfrac{1}{2^2}+\dfrac{2}{2^3}+........+\dfrac{99}{2^{100}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+......+\dfrac{100}{2^{100}}\right)\)
\(\Leftrightarrow B=\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)
Đặt :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+......+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
\(\Leftrightarrow B=1-\dfrac{1}{2^{100}}-\dfrac{100}{2^{100}}\)
\(\Leftrightarrow B=\dfrac{2^{100}-101}{2^{100}}\)
E = 2100 - 299 +298- 297+....+22+2
=> 2A= 2101- 2100+ 299 - 298+......+23 - 22
=> 2A +A = 2101-2100+ 299- 298+......+23+22+ 2100- 299+298 -297+......+22 -2
=> 3A= 22001-2
=> \(\frac{2^{2001}-2}{3}\)
C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự