Tính B= 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+...+20)
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\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+...+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)
\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)
Viết lại đề bài
\(B=1+\frac{1}{2\left(1+2\right)}+\frac{1}{3\left(1+2+3\right)}+\frac{1}{4\left(1+2+3+4\right)}+...+\frac{1}{20\left(1+2+3+4...+20\right)}\)
B=1+12(1+2)+13(1+2+3)+...+120(1+2+...+20)B=1+12(1+2)+13(1+2+3)+...+120(1+2+...+20)
B=1+12.2.3:2+13.3.4:2+...+120.20.21:2B=1+12.2.3:2+13.3.4:2+...+120.20.21:2
B=22+32+...+212B=22+32+...+212
B=2+3+...+212B=2+3+...+212
B=2302B=2302
⇒B=115