Question

[Hoàng Thu Hường]

tính tổng sau:    A=1/12+1/20=1/42+......+1/132.

giải giúp mk nha,mk đang cần gấp

Answer 1

[Mạnh Lê]

\(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{42}\)+ .... + \(\frac{1}{132}\)

= \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ ... + \(\frac{1}{11.12}\)

= \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{11}\)- \(\frac{1}{12}\)

= \(\frac{1}{3}\)- \(\frac{1}{12}\)

= \(\frac{1}{4}\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{11.12}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{3}-\frac{1}{12}\)

\(=\frac{1}{4}\)