Chứng minh rằng: 1/2^2+1/3^2+1/4^2+…+1/N^2<3/4

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1. Đáp án:

    Đặt A=122+132+...+1n2A=122+132+…+1n2

   ⇒A=122+(132+..+1n2)⇒A=122+(132+..+1n2)

   ⇒A=122+B⇒A=122+B

   Xét:BB

   ⇒B=(132+..+1n2)<(12×3+..+1n(n+1⇒B=(132+..+1n2)<(12×3+..+1n(n+1

   ADCT:kn(n+k)=1n−1n+kADCT:kn(n+k)=1n−1n+k

   ⇒B<12−13+..+1n−1n+1⇒B<12−13+..+1n−1n+1

   ⇒B<12−1n+1⇒B<12−1n+1

   Do đó:A=122+B<122+12−1n+1A=122+B<122+12−1n+1

   ⇒A<34−1n+1⇒A<34−1n+1

   Vì 34−1n+1<3434−1n+1<34

   ⇒A<34

    

   Giải thích các bước giải:

    

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2. Đáp án:

    Dưới

   Giải thích các bước giải:

    Đặt $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{n^2}$

   $⇒A=\dfrac{1}{2^2}+(\dfrac{1}{3^2}+..+\dfrac{1}{n^2})$

   $⇒A=\dfrac{1}{2^2}+B$

   Xét:$B$

   $⇒B=(\dfrac{1}{3^2}+..+\dfrac{1}{n^2})<(\dfrac{1}{2×3}+..+\dfrac{1}{n(n+1}$

   $ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$

   $⇒B<\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n+1}$

   $⇒B<\dfrac{1}{2}-\dfrac{1}{n+1}$

   Do đó:$A=\dfrac{1}{2^2}+B<\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{n+1}$

   $⇒A<\dfrac{3}{4}-\dfrac{1}{n+1}$

   Vì $\dfrac{3}{4}-\dfrac{1}{n+1}<\dfrac{3}{4}$

   $⇒A<\dfrac{3}{4}$

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