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a)Ta có:S1=5+52+53+…+599+5100
=>5.S1=52+53+54+…+5100+5101
=>5.S1-S1=52+53+54+…+5100+5101-5-52-53-…-599-5100
=>4.S1=5101-5
=>\(S_1=\frac{5^{101}-5}{4}\)
b)S2=2+22+23+…+299+2100
=>2.S2=22+23+24+…+2100+2101
=>2.S2-S2=22+23+24+…+2100+2101-2-22-23-…-299-2100
=>S2=2101-2
2S1=52+53+54+....+5100+5101
2S1-s1=5101-5
S1=5101-5
b) S2=2101-2
J=6 + 16 + 30 + 48 +...+ 19600 + 19998
Chia cả 2 vế cho 2 ta được
B/2 = 3 + 8 + 15 + 24 + ......... + 98000+ 9999
B/2= 1x3+2x4+3x5+4x6+…….+98x100+99x101
B/2= 100/6[(100-1)x(2x100+1)] = 328350
-> B =328350x2=656700
K=2 + 5 + 9 + 14 + ....+ 4949 + 5049
Nhân cả 2 vế với 2 ta được
2xD=1x4+ 2x5+ 3x6+ 4x7+……..+98x101+99x102
2xD = 1(2+2)+2(3+2)+3(4+2)+...+99(100+2)
2xD = 1x2+1x2+2x3+2x2+3x4+3x2+...+99x100+99x2
2xD= (1x2+2x3+3x4+...+99x100)+2(1+2+3+...+99)
2xD = 333300 + 9900 = 343200
-> D= 343200 :2 =171600
ta có: M = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +......+ 99/3^99 - 100/3^100
=> 3.M = 1 - 2/3 + 3/3^2 - 4/3^3 +.......+ 99/3^98 - 100/3^99
=> 3M + M = ( 1 - 2/3 + 3/3^2 - 4/3^3 +.........+ 99/3^98 - 100/3^99 ) + ( 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +....+ 99/3^99 - 100/3^100 )
=> 4.M = 1- 1/3 + 1/3^2 - 1/3^3 +........+ 1/3^98 - 1/3^99 - 100/3^100
=> 12.M = 3 - 1 + 1/3 - 1/3^2 +.......+ 1/3^97 - 1/3^98 - 1/3^99
=> 12M + 4M = ( 3 - 1 + 1/3 - 1/3^2 +......+ 1/3^97 - 1/3^98 - 1/3^99 ) + ( 1 - 1/3 + 1/3^2 - 1/3^3 +.......+ 1/3^99 - 1/3^100 )
=> 16M = 3 - 101/3^99 - 100/3^100
vù 16M < 3
=> M < 3/16
vậy M < 3/16
tk cho mk nha,mk bị âm rùi
a) \(A=2+2^2+2^3+2^4+.....+2^{98}+2^{99}\)
\(\Rightarrow2A=2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\right)-\left(2+2^2+2^3+2^4+.....+2^{98}+2^{99}\right)\)
\(\Rightarrow A=2^{100}-2\)
b) \(B=2+2^4+2^7+......+2^{97}+2^{100}\)
\(\Rightarrow2^3B=2^4+2^7+......+2^{100}+2^{103}\)
\(\Rightarrow8.B-B=\left(2^4+2^7+......+2^{100}+2^{103}\right)-\left(2+2^4+2^7+......+2^{97}+2^{100}\right)\)
\(\Rightarrow7B=2^{103}-2\)
\(\Rightarrow B=\dfrac{2^{103}-2}{7}\)
\(M=1^2+2^2+3^2+...+99^2+100^2\\ =\dfrac{100\cdot101\cdot201}{6}=338350\)
\(M=1^2+2^2+3^2+...+100^2\\ =1^2+2^2+3^2+...+100^2+1+2+3+...+100-1-2-3-...-100\\ =1^2+1+2^2+2+3^2+3+...+100^2+100-\left(1+2+3+...+100\right)\\ =1\cdot\left(1+1\right)+2\cdot\left(2+1\right)+3\cdot\left(3+1\right)+...+100\cdot\left(100+1\right)-\left(1+2+3+...+100\right)\\ =1\cdot2+2\cdot3+3\cdot4+...+100\cdot101-\left(1+2+3+...+100\right)\)
Đặt \(N=1\cdot2+2\cdot3+3\cdot4+...+100\cdot101\)
\(P=1+2+3+...+100\)
\(3N=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+100\cdot101\cdot3\\ 3N=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+100\cdot101\cdot\left(102-99\right)\\ 3N=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+100\cdot101\cdot102-99\cdot100\cdot101\\ 3N=100\cdot101\cdot102-0\cdot1\cdot2\\ 3N=1030200-0\\ 3N=1030200\\ N=\dfrac{1030200}{3}\\ N=343400\)
\(P=1+2+3+...+100\\ 2P=1+2+3+...+100+1+2+3+...+100\\ 2P=\left(1+100\right)+\left(2+99\right)+\left(3+98\right)+...+\left(100+1\right)\\ 2P=101+101+101+...+101\left(100\text{ số hạng }101\right)\\ 2P=101\cdot100=10100\\ P=\dfrac{10100}{2}=5050\)
\(M=N-P=343400-5050=338350\)