Tính tổng: a, E=1+3+6+...+4950
b, D=2+6+12+...+9900
tính tổng : E=1+3+6+...+4950
D=2+6+12+...+9900
Tính tổng:a, E=1+3+6+...+4950
b, D=2+6+12+...+9900
a)E=1+3+6+...+4950
2E=1.2+3.2+6.2+...+4950.2
2E=2+6+12+...+9900
Ta có: Xét D=1.2+3.2+6.2+...+4950.2
3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3D=99.100.101
D=333300
Thay D vào E ta được 2E=333300 => E=166650
b)B=1+3+6+12+...+9900
2B=1.2+3.2+6.2+12.2+...+9900.2
2B=2+6+12+24+...+19800
Ta có xét A=1.2+3.2+6.2+12.2+...+9900.2
3A=1.2.3+3.2.6-1.2.3+...100.2.3
3A=98.100.102
A=33320
ta thay A vào B; 2B=33320=>B=16660
Tính tổng:a,E=1+3+6+...+4950
b,D=2+6+12+...+9900
Tính tổng D=2+6+12+....+9900
a)Tìm các số a,b,c biết rằng a/2=b/3=c/4 và a+2b+3c=-20
b) Tính tổng: S= 1/2+1/6+1/12+...+1/9900
1.
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}$
$=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1$
$\Rightarrow a=2(-1)=-2; b=3(-1)=-3; c=4(-1)=-4$
2.
$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}$
$=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
a)Tìm các số a,b,c biết rằng a/2=b/3=c/4 và a+2b+3c=-20
b) Tính tổng: S= 1/2+1/6+1/12+...+1/9900
a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)
b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).
Vậy : \(S=\dfrac{99}{100}.\)
a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)
\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)
b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)
1/2 + 1/6 + 1/12 + .......... + 1/9900 tính tổng
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
**** cho mình nha
Tính nhanh tổng sau
A=1/6+1/12+1/20+...1/9900
B=1/3+1/9+1/27+1/81+...=1/2187
Tính tổng
C=(1-1/2)+(1-1/6)+(1-1/12)+......+(1-1/9900)
C = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{9900}\right)\)(99 CẶP)
= \(\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)(99 SỐ HẠNG 1)
= \(1.99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
= \(99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(99-\left(1-\frac{1}{100}\right)\)
= \(99-\frac{99}{100}\)
= \(\frac{9801}{100}\)
Vậy \(C=\frac{9801}{100}\)
Chúc bạn học tốt !!!!!