Toán chứng minh S=1/5^2-1/5^4+1/5^6-…..+1/5^2010-1/5^2012 <1/26 05/07/2021 By Harper chứng minh S=1/5^2-1/5^4+1/5^6-…..+1/5^2010-1/5^2012 <1/26
Ta có : `S=1/5^2-1/5^4+1/5^6-…..+1/5^2010-1/5^2012` $⇔ 5².S = 1 – \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$ $⇔ 5².S + S = \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}\bigg) + \bigg(1- \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$ $⇔ 26S = 1- \dfrac{1}{5^{2014}}$ Mà `1- \frac{1}{5^{2014}} < 1` `⇔ 26S < 1` `⇔ S < 1/26` `⇒ ĐPCM` Học tốt ! Trả lời
Giải thích các bước giải: Ta có : $S = \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}$ $\to 25.S = 1 – \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$ $\to 25.S + S = \bigg( \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}\bigg) + \bigg(1- \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}\bigg)$ $\to 26S = 1- \dfrac{1}{5^{2014}} < 1$ $\to S < \dfrac{1}{26}$ $(đpcm)$ Trả lời
Ta có :
`S=1/5^2-1/5^4+1/5^6-…..+1/5^2010-1/5^2012`
$⇔ 5².S = 1 – \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$
$⇔ 5².S + S = \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}\bigg) + \bigg(1- \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$
$⇔ 26S = 1- \dfrac{1}{5^{2014}}$
Mà `1- \frac{1}{5^{2014}} < 1`
`⇔ 26S < 1`
`⇔ S < 1/26`
`⇒ ĐPCM`
Học tốt !
Giải thích các bước giải:
Ta có : $S = \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}$
$\to 25.S = 1 – \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}$
$\to 25.S + S = \bigg( \dfrac{1}{5^2}-\dfrac{1}{5^4}+\dfrac{1}{5^6}-….+\dfrac{1}{5^{2010}} – \dfrac{1}{5^{2012}}\bigg) + \bigg(1- \dfrac{1}{5^2}+\dfrac{1}{5^4}-\dfrac{1}{5^6}+….+\dfrac{1}{5^{2012}} – \dfrac{1}{5^{2014}}\bigg)$
$\to 26S = 1- \dfrac{1}{5^{2014}} < 1$
$\to S < \dfrac{1}{26}$ $(đpcm)$