Math, asked by ItZzMissKhushi, 3 months ago

Solve :—
(1 -  \frac{1}{3} )(1 -  \frac{1}{4})(1 -  \frac{1}{5})(1 -  \frac{1}{6}).. \: .. \: .(1 -  \frac{1}{13})

Answers

Answered by ItZzKhushi
3

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Solve:—

(1 - \frac{1}{3} )(1 - \frac{1}{4})(1 - \frac{1}{5})(1 - \frac{1}{6}).. \: .. \: .(1 - \frac{1}{13})

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 = (1 -  \frac{1}{3} )(1 -  \frac{1}{4})(1 -  \frac{1}{5}  )(1 -  \frac{1}{6} ).. \: ..(1 -  \frac{1}{13} )

 = ( \frac{1}{1}   - \frac{1}{3} )( \frac{1}{1}  -  \frac{1}{4} )( \frac{1}{1}   - \frac{1}{5} )( \frac{1}{1}   - \frac{1}{6}).. \: .. \: ..( \frac{1}{1}   - \frac{1}{13} )

 =( \frac{3 - 1}{3} )( \frac{4 - 1}{4} )( \frac{5 - 1}{5} )( \frac{6 - 1}{6} ).. \: .. \: ..( \frac{13 - 1}{13} )

 =  \frac{2}{ \cancel3}  \times   \frac{\cancel3}{\cancel4} \times  \frac{\cancel4}{\cancel5}   \times  \frac{\cancel5}{\cancel6}  \times  \frac{\cancel6}{\cancel7}  \times  \frac{\cancel{7}}{\cancel8}  \times  \frac{\cancel8}{\cancel9}  \times  \frac{\cancel9}{\cancel10}  \times  \frac{\cancel{10}}{\cancel{11}}  \times  \frac{\cancel{11}}{\cancel{12}}  \times  \frac{\cancel{12}}{13}

 =  \frac{2}{13}

Answered by Itzkirtihere
4

SOLVE→_→

(1 -  \frac{1}{3} )(1 -  \frac{1}{4} )(1 -  \frac{1}{5} ) \\ (1 -  \frac{1}{6} )(1 -  \frac{1}{3} )

ANSWER→_→

  = (1 -  \frac{1}{3} )(1 -  \frac{1}{4} )(1 -  \frac{1}{5} )(1 -  \frac{1}{6} )(1 -  \frac{1}{13} )

 = ( \frac{1}{1}  -  \frac{1}{3} )( \frac{1}{1}  -  \frac{1}{4} )(  \frac{1}{1}  -  \frac{1}{5} )( \frac{1}{1}  -  \frac{1}{6} )( \frac{1}{1}  -  \frac{1}{13} )

 = ( \frac{3 - 1}{3} )( \frac{4 - 1}{4} )( \frac{5 - 1}{5} )( \frac{6 - 1}{6} )( \frac{13 - 1}{13} )

 =  \frac{2}{13}

Hope it's helpful for u.

have a great day ☺️

Hello kushi sis

itz kirti here new I'd

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