Question

cho M =1+3+3²+3³+......+3¹¹⁸+3¹¹⁹

N=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+......+\(\dfrac{1}{2009^2}\)+\(\dfrac{1}{2010^2}\)

CHỨNG TỎ RẰNG

a)M\(⋮\)13

b)N<1

Answer 1

a) M =1+3+3²+3³+......+3¹¹⁸+3¹¹⁹
M = ( 1+3+3² ) +...+ ( 3¹¹⁷ + 3¹¹⁸+3¹¹⁹ )
M = 1. ( 1+3+3² ) + ... + 3¹¹⁷ . ( 3¹¹⁷ + 3¹¹⁸+3¹¹⁹ )
M = ( 1+3+3² ) .( 1 + ... + 3¹¹⁷ )
M = 13 . ( 1 + ... + 3¹¹⁷ ) \(⋮\) 13 (đpcm )

Answer 2

b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)

=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)

Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1

Answer 3