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\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4.4...100.101}\)
\(=\frac{\left(1.2.3...100\right)\left(1.2.3...100\right)}{\left(1.2.3..100\right)\left(2.3.4...101\right)}=\frac{1}{101}\)
\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(P=1+\frac{1}{2}.3+\frac{1}{3}.6+\frac{1}{4}.10+....+\frac{1}{2016}.2033136\)
\(P=1+\frac{3}{2}+4+\frac{5}{2}+....+\frac{2017}{2}\)
\(P=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2017}{2}\)
\(P=\frac{2+3+4+5+....+2017}{2}=\frac{2035152}{2}=1017576\)
Bài làm:
Ta có:
\(B=-66\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(B=\left(-66\right).\frac{1}{2}+66.\frac{1}{3}-66.\frac{1}{11}-124.\left(37+63\right)\)
\(B=-33+22-6-124.100\)
\(B=17-12400\)
\(B=-12383\)
\(B=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(=-66.\frac{1}{2}-\left(-66\right).\frac{1}{3}+\left(-66\right).\frac{1}{11}+\left(-124\right).37+63.\left(-124\right)\)
\(=-33+22-6+\left(-124\right).\left(37+63\right)\)
\(=-11-6+\left(-124\right).100\)
\(=-17-12400\)
\(=-12417\)
TA CÓ: \(1+2+2^2+2^3+...+2^{100}\)
\(=2^0+2^1+2^2+2^3+...+2^{100}\)
= TO LẮM BẠN