Tính nhanh B= 1+½.(1+2)+⅓.(1+2+3)+¼.(1+2+3+4)+...+1/20.(1+2+3+...+20)
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\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+...+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)
\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)
`1/2+2/4+3/6+4/8+5/10+6/12`
`=1/2+1/2+1/2+1/2+1/2+1/2`
`=1/2*6=3`
`1/3+1/4+1/5+8/10+20/15+20/30`
`=(1/3+1/4)+(1/5+4/5)+(4/3+2/3)`
`=7/12+1+2`
`=7/12+3=43/12`
\(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{6}+\dfrac{4}{8}+\dfrac{5}{10}+\dfrac{6}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{1}{2}\times6=3\)
\(------\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{8}{10}+\dfrac{20}{15}+\dfrac{20}{30}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{4}{3}+\dfrac{2}{3}\)
\(=\left(\dfrac{1}{3}+\dfrac{4}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\dfrac{1}{4}\)
\(=\dfrac{7}{3}+1+\dfrac{1}{4}\)
\(=\dfrac{28}{12}+\dfrac{12}{12}+\dfrac{3}{12}\)
\(=\dfrac{43}{12}\)
Xét số hạng tổng quát thứ n (n nguyên và n>1), ta có
1/n(1+2+...+n)=[n(n+1)/2]/n= [n(n+1)]/(2n)
Do đó
B = 1 + 1/2 (1 + 2) + 1/3 (1 + 2 + 3) + 1/4 (1 + 2 + 3 +4) + ...+ 1/20 (1 + 2 +... + 20)
=1 +[2(2+1)]/(2.2) +[3(3+1)]/(2.3) +[4(4+1)]/(2.4) +... +[20(20+1)]/(2.20)
=1+3/2 +4/2 +5/2 +... +21/2
=(2+3+4+5+...+20)/2=104,5 . TICH CHON MINH NHA CAC BAN THI CA NAM SE GAP NHIEU DIEU MAY MAN DAY
Xét số hạng tổng quát thứ n (n nguyên và n>1), ta có
1/n(1+2+...+n)=[n(n+1)/2]/n= [n(n+1)]/(2n)
Do đó
B = 1 + 1/2 (1 + 2) + 1/3 (1 + 2 + 3) + 1/4 (1 + 2 + 3 +4) + ...+ 1/20 (1 + 2 +... + 20)
=1 +[2(2+1)]/(2.2) +[3(3+1)]/(2.3) +[4(4+1)]/(2.4) +... +[20(20+1)]/(2.20)
=1+3/2 +4/2 +5/2 +... +21/2
=(2+3+4+5+...+20)/2=104,5