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\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{100}\left(1+2+3+....+100\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{100}.\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{101}{2}\)
\(=\frac{2+3+4+....+101}{2}\)
\(=\frac{\frac{101.102}{2}-1}{2}\)
\(=2575\)
Vậy \(S=2575\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{100}\left(1+2+3+....+100\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+\frac{1}{4}.\frac{4\left(4+1\right)}{2}+.....+\frac{1}{100}.\frac{100\left(100+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{100+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{101}{2}\)
\(=\frac{2+3+4+....+101}{2}\)
\(=\frac{\frac{101\left(101+1\right)}{2}-1}{2}=5150.5\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
=> S = \(1-\frac{1}{2^{100}}\)
Tính :
\(S=2+2^2+2^3+...+2^{100}\)
\(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
a)S=2+22+23+...+2100
2S=2(2+22+23+...+2100)
2S=22+23+...+2101
2S-S=(22+23+...+2101)-(2+22+23+...+2100)
S=2101-2
b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)
\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2P=1-\frac{1}{3^{100}}\)
\(P=\left(1-\frac{1}{3^{100}}\right):2\)
Xét : \(\frac{1}{100}-\frac{1}{n^2}=\frac{n^2-100}{100n^2}=\frac{\left(n-10\right)\left(n+10\right)}{100n^2}\)
Áp dụng , đặt biểu thức cần tính là A , ta có :
\(A=\left(\frac{1}{100}-\frac{1}{1^2}\right)\left(\frac{1}{100}-\frac{1}{2^2}\right)\left(\frac{1}{100}-\frac{1}{3^2}\right)...\left(\frac{1}{100}-\frac{1}{20^2}\right)\)
\(=\frac{\left(1-10\right)\left(1+10\right)}{100.1^2}.\frac{\left(2-10\right)\left(2+10\right)}{100.2^2}.\frac{\left(3-10\right)\left(3+10\right)}{100.3^2}...\frac{\left(10-10\right)\left(10+10\right)}{100.10^2}...\frac{\left(20-10\right)\left(20+10\right)}{100.20^2}\)
Nhận thấy trong A có một nhân tử (10-10) = 0 nên A = 0
làm thế thì hơi dài đấy Hoàng Lê Bảo Ngọc
ta nhận thấy trong biểu thức chứa thừa số \(\frac{1}{100}-\left(\frac{1}{10}\right)^2=\frac{1}{100}-\frac{1}{100}=0\)
=>biểu thức ấy =0
Lời giải:
\(S=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+...+\frac{100.101}{2.100}\\ =1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{101}{2}\\ =\frac{2+3+4+...101}{2}=\frac{1+2+3+...+101}{2}-\frac{1}{2}\\ =\frac{101.102}{2.2}-\frac{1}{2}=2575\)