Tính a(a+1)(a+2)(a+3)...(a+n) với a>0
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Ta có : E = 1 + 3 + 32 + 33 + ..... + 390
=> 3E = 3 + 32 + 33 + ..... + 391
=> 3E - E = 391 - 1
=> 2E = 391 - 1
=> \(E=\frac{3^{91}-1}{2}\)
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
c) \(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}=1-\frac{1}{97}=\frac{96}{97}\)
1, Ta có a^3+b^3+c^3=3abc
-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2
-> (a+b)3 + c^3 - 3ab(a+b+c)=0
-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0
-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0
Th1: a+b+c=0
->P= a+b/2 . b+c/2 . c+a/2
= (-c)(-a)(-b)/2=-1
TH2 a^2+b^2+c^2-ab-bc-ca=0
->2a^2+2b^2+2c^2-2ab-abc-2ac=0
->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
-> (a-b)^2+(a-c)^2+(b-c)^2=0
Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0
Dấu = xảy ra (=)a-b=0
b-c=0
a-c=0
-> a=b=c
->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8
\(\lim\dfrac{1+a+...+a^n}{1+b+...+b^n}=\lim\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}=\lim\dfrac{\left(1-a^n\right)\left(1-b\right)}{\left(1-b^n\right)\left(1-a\right)}=\dfrac{1-b}{1-a}\)
\(\Rightarrow\dfrac{1-b}{1-a}=\dfrac{2}{3}\Leftrightarrow3-3b=2-2a\)
\(\Leftrightarrow2a-3b=-1\)