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S = (30/2 + 1/2) + (31/2 + 1/2) + (32/2 + 1/2) + (33/2 + 1/2) +..+ 3n-1/2 + 1/2
S = n.(1/2) + (1/2)[3^0 + 3^1 + 32 +...+ 3n-1]
S = n/2 + (3^n - 1)/4 = (3^n + 2n - 1)/4
S = (30/2 + 1/2) + (31/2 + 1/2) + (3²/2 + 1/2) + (3³/2 + 1/2) +..+ 3(n-1)/2 + 1/2
S = n.(1/2) + (1/2)[30 + 31 + 3² +...+ 3(n-1)]
S = n/2 + (3n - 1)/4 = (3n + 2n - 1)/4
\(1+a^2+a^4+a^6+.....+a^{2n}\)
\(\Rightarrow a^2.S1=a^2+a^4+a^6+a^8+.....+a^{2\left(1+n\right)}\)
\(\Rightarrow a^2.S1-S1=\left(a^2+a^4+....+2^{2\left(1+n\right)}\right)-\left(1+a^2+a^4+....+2^{2n}\right)\)
\(\Rightarrow S1\left(a-1\right)\left(a+1\right)=a^{2\left(1+n\right)}-1\)
\(\Rightarrow S1=\frac{a^{2\left(1+n\right)}-1}{\left(a-1\right)\left(a+1\right)}\)
a: \(s1=\dfrac{999\cdot\left(999+1\right)}{2}=499500\)
b: =>n(n+1)/2=378
=>n(n+1)=756
=>n^2+n-756=0
=>n=27
c: \(5Q=5+5^2+...+5^{101}\)
=>\(4\cdot Q=5^{101}-1\)
hay \(Q=\dfrac{5^{101}-1}{4}\)
Bài 1
a) {42-[52-(92-16×5)30×8×3]3-14}
= {42-[52-(81 - 80 )30×8×3]3-14}
= {42-[52- 1 30×8×3]3-14}
= {42-[52- 1 ×8×3]3-14}
= {42-[52- 24 ]3-14}
= {42-[25 - 24 ]3-14}
= 42- 1 3-14
= 42- 1 -14
= 16 -1-14
= 1
b,
b) 568-{5[143-(22-1)2]+10}: 10
= 568-{5[143-(4 -1)2]+10}: 10
= 568-{5[143-32]+10}: 10
= 568-{5[143-9 ]+10}: 10
= 568-{5. 134+10}: 10
= 568-{ 670 +10}: 10
= 568- 680: 10
= 568- 68
= 500
bài 2
a,
a) 20105 × ( X - 60 ) = 20106
X - 60 = 20106 : 20105
X - 60 = 20101
X - 60 = 2010
X = 2010 + 60
X= 2070
b) 80 - ( 4×52 - 3×23)= 210 - ( x - 4 )
80 - ( 4×25 - 3×8)= 210 - ( x - 4 )
80 - ( 100 - 24)= 210 - ( x - 4 )
80 - 76 = 210 - ( x - 4 )
4 = 210 - ( x - 4 )
hay 210 - ( x - 4 ) = 4
1024 - ( x-4 ) = 4
x-4 = 1024 -4
x-4 = 1020
x = 1020+4
x= 1024
bài 3
a) S= 1+2+3+4+....+1000 ( 1 000 số hạng )
S= ( 1000 + 1 ) * 1 000 : 2
S= 500 500
b)S=1+3+5+....+2003 ( 1 002 số hạng )
S = ( 2003 +1 ) * 1 002 : 2
S= 1 004 004
c)S=1+2+3+...+2013 ( 2013 số hạng )
S = ( 2013 + 1 ) * 2013 : 2
S= 2 027 091
d)S= 3+6+9+..+2010 ( 670 số hạng )
S= ( 2010+3 ) * 670 : 2
S= 674 355
P/s : Mệt
Sn+1−Sn=(n+1)3=n3+3.n2+3.n+1Sn=CSn=An4+B.n3+C.n2+D.nSn+1−Sn=A.(n+1)4+B.(n+1)3+C.(n+1)2+D.(n+1)−An4−B.n3−C.n2−D.n=n3+3.n2+3.n+14.A.n3+(6A+3B).n2+(4A+3B+2C)n+(A+B+C+D)=n3+3.n2+3.n+1A=14,B=12,C=14,D=0Sn=C+1/4.n4+1/2.n3+1/4.n2S1=1⇒C=0Sn=14.n2.(n+1)2Sn+1−Sn=(n+1)3=n3+3.n2+3.n+1Sn=CSn=An4+B.n3+C.n2+D.nSn+1−Sn=A.(n+1)4+B.(n+1)3+C.(n+1)2+D.(n+1)−An4−B.n3−C.n2−D.n=n3+3.n2+3.n+14.A.n3+(6A+3B).n2+(4A+3B+2C)n+(A+B+C+D)=n3+3.n2+3.n+1A=14,B=12,C=14,D=0Sn=C+1/4.n4+1/2.n3+1/4.n2S1=1⇒C=0Sn=14.n2.(n+1)2
chốt đáp án
Sn=n2.(n+1)24