Chứng tỏ rằng: $\frac{1}{3^{2} }$ + $\frac{1}{5^{2} }$ + $\frac{1}{7^{2} }$ +…+ $\frac{1}{2021^{2} }$ < $\frac{1}{12}$
Chứng tỏ rằng: $\frac{1}{3^{2} }$ + $\frac{1}{5^{2} }$ + $\frac{1}{7^{2} }$ +…+ $\frac{1}{2021^{2} }$ < $\frac{1}{12}$
Đáp án:
1/3^2+1/5^2+1/7^2+…+1/(2n+1)^2 < 1/1.3+1/3.5+1/5.7+…+1/(2n-1)(2n+1)
= 1/2(1-1/3+1/3-1/5+1/5-1/7+…+1/(2n-1)-1/(2n+1)
= 1/2(1-1/(2n+1))
= 1/2 . 2n/(2n+1)
= 2n/2(2n+1)