3 replies to this topic

[tunglubelu]

a) CMR:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$
b) Giả sử; $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1334}+\frac{1}{1335}=\frac{p}{q}$ với p, q $\in N*$, (p,q)=1
CM: P chia hết cho 2013

Edited by dark templar, 18-03-2013 - 17:58.

[Christian Goldbach]

a) CMR:
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$
b) Giả sử; $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1334}+\frac{1}{1335}=\frac{p}{q}$ với p, q $\in N*$, (p,q)=1
CM: P chia hết cho 2013

Ta đặt A=$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n} $ \Rightarrow A = A+2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n} = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$

Edited by MrMathCSKH0110, 18-03-2013 - 17:49.

[Jayce Tran]

Ta đặt A=$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n} $ \Rightarrow A = A+2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n} = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$

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[Christian Goldbach]

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