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a,1-3+5-7+9-.......+33-35
=(1+5+9+....+33)-(3+7+11+...+35)
=153-171
=-18
Tick mk vài cái lên 300 mk giải nốt phần b
b) 32, 33, 34 , 35.
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
Ta có: 3A = 3.(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−1
⇒ A = 3101−1
2
Vậy A = 3101−1
2
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
Tham khảo
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
a) Đặt \(A'=3^2+3^3+...+3^{100}\), ta có \(A=3-A'\)
Ta tính A'.
\(3A'=3^3+3^4+...+3^{100}+3^{101}\)
\(A'=3^2+3^3+...+3^{100}\)
\(\Rightarrow2A'=3^{101}-9\Rightarrow A'=\frac{3^{101}-9}{2}\)
Vậy \(A=3-\frac{3^{101}-9}{2}=\frac{6-3^{101}+9}{2}=\frac{15-3^{101}}{2}\)
b) Đặt \(C=4+4^3+4^5+...+4^{101}\)
\(D=1+4^2+4^4+4^6+...+4^{100}\)
Ta có \(16C=4^3+4^5+4^7+...+4^{101}+4^{103}\)
\(\Rightarrow15C=4^{103}-4\Rightarrow C=\frac{4^{103}-4}{15}\)
Ta có \(16D=4^2+4^4+4^6+4^8+...+4^{100}+4^{102}\)
\(\Rightarrow15D=4^{102}-1\Rightarrow D=\frac{4^{102}-1}{15}\)
Vậy \(B=-C+D=-\frac{4^{103}-4}{15}+\frac{4^{102}-1}{15}=\frac{4^{102}-4^{103}+3}{15}\)
\(=\frac{3.4^{102}+3}{15}=\frac{4^{102}+1}{5}\)