Question

[Phạm Hồ Hữu Trí]

Cho mik hoi nha, ai nhanh va dung mik se tick.

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

Chung minh A<2

Cam on nha

Answer 1

[tth_new]

Áp dụng công thức \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\). Ta có:

\(\frac{1}{2^2}< 2-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)

.  .  .   .  .

\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)

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\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2-\frac{1}{50}=\frac{99}{50}\)

Vậy:A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2^{\left(đpcm\right)}\)

Answer 2

[giahuy356]

hay thế