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Problem #959

Cm=quy nap \(1+ \dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n-1}<n\)

Added by Nguyễn Lê Ngọc Tùng 11C2 [G] about 1 year ago. Updated about 1 year ago.

Status: Closed Start Date: 30-12-2011
Priority: Normal Due date:
Assigned to: tanphu % Done:


Category: Quy nạp
Target version: -
Votes: 0/0


Cm=quy nap: \(1+ \dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n-1}<n\)
mọi người giúp em... mai em thi rồi!


Updated by tanphu about 1 year ago

  • Subject changed from Cm=quy nap: 1+ \dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n-1}<n to Cm=quy nap \(1+ \dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n-1}<n\)
  • Category set to Quy nạp
  • Status changed from New to Assigned
  • Assigned to set to tanphu

Updated by tanphu about 1 year ago

    Em xem lại đề có trục trặc không? Thầy chưa thấy quy luật trong tổng ở vế trái.

    Updated by icy about 1 year ago

      1/1, 1/2, ..., 1/k, 1/(k+1),... , 1/(2^n-1)

      Updated by tanphu about 1 year ago

      • Status changed from Assigned to Closed