áp dụng định nghĩa giới hạn của dãy số, chứng tỏ : $lim \frac{1+2+...+n}{n^{2}}
=\frac{1}{2}$
nhấn V với mũi tên xanh chỉ lên cho mk đi –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:50 PM

$1+2+3+...+n=\frac{n(n+1)}{2}\\=>\mathop {\lim }\limits\frac{1+2+3+...+n}{n^2}\\=\lim\frac{n+1}{2n}\\=\lim \frac{1+\frac{1}{n}}{2}\\=\frac{1}{2}\\=>DPCM$

cái đâu với cuốn là n cộng 1 á –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:49 PM
cộng lại thì có 2 lần tổng = n(n cộng 1 ) –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:49 PM
n 1 n ... 4 3 2 1 –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:48 PM
1 2 3 4 ... n n 1 –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:48 PM
viết ra để cộng giống toán cấp 1 á. –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 27-10-14 11:47 PM
cho m hỏi tại sao 1 2 3 ... n =( n(n 1))/2 –  ngongiothansaublabla 27-10-14 11:43 PM

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