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\(A=5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+...+100}\)
A = \(5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+..+100}\)
\(=5x\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=5x\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=2x5x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=10x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{100x101}\right)\)
\(=10x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=10x\left(1-\frac{1}{101}\right)\)
\(=10x\frac{100}{101}\)
\(=\frac{1000}{101}\)
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
Từ 1 đến 100 có Số số hạng là :
( 100 -1 ) : 1 + 1 = 100 ( số )
=> TỪ 1/1 -> 100/100 có 100 số
TA có 1/1 + 2/2 + 3/3 + ... + 99/99 + 100/100
= 1 + 1 + 1 + ... + 1
=> có 100 số 1
= 100 x 1 = 100
Các bạn t i ck mk nếu đúng nha !
a) \(\frac{3}{5}+25-\frac{1}{5}=\left(\frac{3}{5}-\frac{1}{5}\right)+25=\frac{2}{5}+25=\frac{2}{5}+\frac{125}{5}=\frac{127}{5}\)
b) \(13\times3\times32,27+67,63\times39=39\times32,27+67,63\times39\)
\(=39\times\left(32,27+67,63\right)\)
\(=39\times99,9=3196,8\)
(Bạn xem lại đề nhé)
c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{99}{100}\)
\(=\frac{1\times2\times3\times...\times99}{2\times3\times4\times....\times100}=\frac{1}{100}\)
a, \(\frac{3}{5}+25-\frac{1}{5}=\frac{127}{5}\)
b, \(\text{13 x 3 x 32,27 + 67,63 x 39 =}3896,1\)
............................
thank for watching me do homework
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)
\(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)
\(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)
\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6.\left(1-\frac{1}{101}\right)\)
\(=6.\frac{100}{101}=\frac{600}{101}\)
Vậy \(A=\frac{600}{101}\)
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)
\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)
Vay A = ........