$\dfrac{1}{15}$ + $\dfrac{1}{35}$ + …+ $\dfrac{1}{2499}$ 24/08/2021 Bởi Katherine $\dfrac{1}{15}$ + $\dfrac{1}{35}$ + …+ $\dfrac{1}{2499}$
$\frac{1}{15}$ + $\frac{1}{35}$ + … + $\frac{1}{2499}$ = $\frac{1}{2}$ . ($\frac{2}{3.5}$ + $\frac{2}{5.7}$ + … + $\frac{2}{49.51}$) = $\frac{1}{2}$ . (($\frac{1}{3}$ – $\frac{1}{5}$) + ($\frac{1}{5}$ – $\frac{1}{7}$) + … + ($\frac{1}{49}$ – $\frac{1}{51}$)) = $\frac{1}{2}$ . ($\frac{1}{3}$ – $\frac{1}{5}$ + $\frac{1}{5}$ – $\frac{1}{7}$ + … + $\frac{1}{49}$ – $\frac{1}{51}$) = $\frac{1}{2}$ . ($\frac{1}{3}$ – $\frac{1}{51}$) = $\frac{8}{51}$ Chúc học tốt!!! Bình luận
Đáp án: Giải thích các bước giải: Ta có =$\frac{1}{15}$+$\frac{1}{35}$ +…………+$\frac{1}{2499}$ =$\frac{1}{3.5}$ +$\frac{1}{5.7}$ +…………..+$\frac{1}{49.51}$ =$\frac{1}{2}$ .($\frac{1}{3}$ -$\frac{1}{5}$ +$\frac{1}{5}$ -$\frac{1}{7}$ +……..+$\frac{1}{49}$ -$\frac{1}{51}$ ⇒ A=$\frac{1}{2}$ .($\frac{1}{3}$ -$\frac{1}{51}$ ) =8/51.1/2 =8/51 Bình luận
$\frac{1}{15}$ + $\frac{1}{35}$ + … + $\frac{1}{2499}$
= $\frac{1}{2}$ . ($\frac{2}{3.5}$ + $\frac{2}{5.7}$ + … + $\frac{2}{49.51}$)
= $\frac{1}{2}$ . (($\frac{1}{3}$ – $\frac{1}{5}$) + ($\frac{1}{5}$ – $\frac{1}{7}$) + … + ($\frac{1}{49}$ – $\frac{1}{51}$))
= $\frac{1}{2}$ . ($\frac{1}{3}$ – $\frac{1}{5}$ + $\frac{1}{5}$ – $\frac{1}{7}$ + … + $\frac{1}{49}$ – $\frac{1}{51}$)
= $\frac{1}{2}$ . ($\frac{1}{3}$ – $\frac{1}{51}$)
= $\frac{8}{51}$
Chúc học tốt!!!
Đáp án:
Giải thích các bước giải:
Ta có =$\frac{1}{15}$+$\frac{1}{35}$ +…………+$\frac{1}{2499}$
=$\frac{1}{3.5}$ +$\frac{1}{5.7}$ +…………..+$\frac{1}{49.51}$
=$\frac{1}{2}$ .($\frac{1}{3}$ -$\frac{1}{5}$ +$\frac{1}{5}$ -$\frac{1}{7}$ +……..+$\frac{1}{49}$ -$\frac{1}{51}$
⇒ A=$\frac{1}{2}$ .($\frac{1}{3}$ -$\frac{1}{51}$ )
=8/51.1/2
=8/51