Lời giải:
\(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}....+\frac{1}{50.51}\)
Mà:
\(\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}....+\frac{1}{50.51}=\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{50}-\frac{1}{51}=\frac{7}{12}-\frac{1}{51}>\frac{1}{2}\)
Do đó: $B>\frac{1}{2}$
Ta có đpcm.
Lời giải:
B=122+132+...+1502>14+13.4+14.5....+150.51B=122+132+...+1502>14+13.4+14.5....+150.51
Mà:
14+13.4+14.5....+150.51=14+13−14+....+150−151=712−151>1214+13.4+14.5....+150.51=14+13−14+....+150−151=712−151>12
Do đó: B>12B>12
Ta có đpcm.
