Question

tính \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{49.54}\right).\dfrac{1-3-...-51}{108}\)

Answer 1

`S = (1/(4.9) + 1/(9.14) + ... +1/(49.54)) . (1-3-...-51)/108`

Đặt: `{(A = 1/(4.9) + 1/(9.14) + ... +1/(49.54)),(B = (1-3-...-51)/108):}`

Ta có:

`-> A = 1/(4.9) + 1/(9.14) + ... +1/(49.54)`

`5A = 5/(4.9) + 5/(9.14) + ... +5/(49.54)`

`5A = 1/4 - 1/9 + 1/9 - 1/14 + ... + 1/49 - 1/54`

`5A = 1/4 - 1/54`

`5A = 25/108`

`A = 5/108`

`-> B = (1-3-...-51)/108`

`B = (1 - (3+7+...+51))/108`

`B = (1 - (51 + 3) . [(51-3) : 2 + 1] : 2)/108`

`B = (1 - 54 . [48 : 2 + 1] : 2)/108`

`B = (1 - 54 .25 : 2)/108`

`B = (1 - 675)/108`

`B = (-674)/108`

`B = (-337)/54`

`=> S = A . B =  5/108 . (-337)/54 = -1685/5832`