chứng minh rằng : a) $\frac{a}{n(n+a)}$ = $\frac{1}{n}$ – $\frac{1}{n+a}$ ( n,a∈N* )
b) áp dụng câu a , tính :
A = $\frac{1}{15}$ + $\frac{1}{35}$ + … + $\frac{1}{2499}$
B = $\frac{3}{5.8}$ + $\frac{11}{8.19}$ + $\frac{12}{19.31}$ + $\frac{70}{31.101}$ + $\frac{99}{101.200}$
$a$) Ta có:
`VP = 1/n – 1/{n+a} = {(n+a)}/{n.(n+a)} – n/{n.(n+a)} = {n+a-n}/{n.(n+a)} = a/{n.(n+a)} = VT` ($đ.p.c.m$)
$b$)
`A = 1/15 + 1/35 + …. + 1/2499`
`⇔ A = 1/{3.5} + 1/{5.7} + ….. + 1/{49.51}`
`⇔ 2A = 2/{3.5} + 2/{5.7} + …. + 2/{49.51}`
`⇔ 2A = 1/3 – 1/5 + 1/5 – 1/7 + … + 1/49 – 1/51`
`⇔ 2A = 1/3 – 1/51`
`⇔ 2A = 16/51`
`⇔ A = 8/51`
`B = 3/{5.8} + 11/{8.19} + 12/{19.31} + {70}/{31.101} + 99/{101.200}`
`⇔ B = 1/5 – 1/8 + 1/8 – 1/19 + 1/19 – 1/31 + 1/31 – 1/101 – 1/101 – 1/200`
`⇔ B = 1/5 – 1/200`
`⇔ B = 39/200`
a, a/n(n+a) = 1/n – 1/n+a = n + a – n / n(n+a) = a / n(n+a) ( đpcm )
b, A = 1/15 + 1/35 + … + 1/2499
= 1 / 3.5 + 1/5.7 + … + 1/49.51
= 1/2 ( 1/3 – 1/5 + 1/5 – 1/7 + … + 1/49 – 1/51 )
= 1/2 ( 1/3 – 1/51 ) = 1/2 . 8/51 = 8 / 102 = 4/51