$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + $\frac{1}{15}$ + … + $\frac{1}{x(2x+1)}$ = $\frac{1}{10}$ x ∈ N*
$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + $\frac{1}{15}$ + … + $\frac{1}{x(2x+1)}$ = $\frac{1}{10}$ x ∈ N*
Đáp án:
Ta có :
A = $\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + … + $\frac{1}{x(2x+1)}$ = $\frac{1}{10}$
=> A = $\frac{2}{6}$ + $\frac{2}{12}$ + $\frac{2}{20}$ + … + $\frac{2}{2x(2x+1)}$ = $\frac{1}{10}$
=> A = 2.($\frac{1}{6}$ + $\frac{1}{12}$ + $\frac{1}{20}$ + … + $\frac{1}{2x(2x+1) })$ = $\frac{1}{10}$
=> A = 2 . ( $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + … + $\frac{1}{x(2x+1)})$ = $\frac{1}{10}$
=> A = 2. ( $\frac{1}{2}$ – $\frac{1}{3}$ +$\frac{1}{3}$ – $\frac{1}{4}$ + $\frac{1}{4}$ -$\frac{1}{5}$ + … +$\frac{1}{2x}$ – $\frac{1}{2x+1})$ = $\frac{1}{10}$
=> A = 2.($\frac{1}{2}$ – $\frac{1}{2x+1})$ ) = $\frac{1}{10}$
=> A = 2.$\frac{2x-1}{2(2x+1)}$ = $\frac{1}{10}$
=> A = $\frac{2x-1}{2x+1}$ = $\frac{1}{10}$
=> 2x – 1 = 1 => x = 1
=> $\frac{2x-1}{2x+1}$ = $\frac{1}{3}$ $\neq$ $\frac{1}{10}$
=> không có giá trị x thỏa mãn
Giải thích các bước giải:
`⇒ 2/6 + 2/12 + … + 2/x(x+1) = 1/10`
`⇒ 2/2.3 + 2/3.4 + … + 2/x(x+1) = 1/10`
`⇒ 2.( 1/2.3 + 1/3.4 + …+ 1/x(x+1)= 1/10 `
`⇒ 1/2 – 1/3 + 1/3 – 1/4 + … + 1/x – 1/x+1 = 1/10 : 2 = 1/20`
`⇒ 1/2 – 1/x+1 = 1/20`
`⇒ 1/x+1 = 1/2 – 1/20 = 9/20`
Vậy ko có gt x nào thỏa mãn