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Đặt an=1+2+3+4+5+...+n
=>an=\(\frac{n\left(n+1\right)}{2}\)
Ta có: Sn=(1+2+3+4+...+n)+(an+1)+(an+2)+(an+3)+...+(an+n)
=an+n.an+an
=an(n+2)
=\(\frac{n\left(n+1\right)}{2}\left(n+2\right)\)
=\(\frac{n\left(n+1\right)\left(n+2\right)}{2}\)
Vậy S100=\(\frac{100.101.102}{2}=515100\)
\(S_2=10+12+14+...+210=\frac{\left(210-10\right)}{2}\cdot\left(210+10\right):2=\frac{200\cdot220}{4}=100\cdot110=11000.\)
\(S_3=21+23+25+...+101=\frac{101-21}{2}\cdot\left(101+21\right):2=\frac{80\cdot122}{4}=40\cdot61=2440\)
v.v
Các bài sau bn cứ tính theo công thức: tổng dãy có quy luật=(số đầu - số cuối) : 2 x (số đầu + số cuối) :2
học tốt ^_^
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
Bằng 515100