B =\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...\frac{2}{97.100}\)
=2.(\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\))
3B=2.(\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\))
3B=2.(\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\))
3B=2.(1-\(\frac{1}{100}\))
3B=2.\(\frac{99}{100}\)=\(\frac{99}{50}\)
=>B=\(\frac{99}{50}:3\)=\(\frac{33}{50}\)
Tick mik nha