Cho $P$=$\frac{1}{2}$ +$\frac{1}{3}$ +…+$\frac{1}{2019}$ và $Q$=$\frac{1}{2018}$ +$\frac{2}{2017}$ +$\frac{3}{2016}$ +…+$\frac{2016}{3}$ +$\frac{2017}{2}$ +$\frac{2018}{1}$ Tính $(Q-2019.P)^{2019}$
Cho $P$=$\frac{1}{2}$ +$\frac{1}{3}$ +…+$\frac{1}{2019}$ và $Q$=$\frac{1}{2018}$ +$\frac{2}{2017}$ +$\frac{3}{2016}$ +…+$\frac{2016}{3}$ +$\frac{
By Athena
Giải thích các bước giải:
Ta có:
$2019.P$ $=$ $2019$ $.$ $(\frac{1}{2} + \frac{1}{3} + $…$ + \frac{1}{2019})$ $=$ $\frac{2019}{2}$ + $\frac{2019}{3}$ +…+ $\frac{2019}{2019}$
$Q$ $=$ $\frac{1}{2018}$ + $\frac{2}{2017}$ + $\frac{3}{2016}$ + … + $\frac{2016}{3}$ + $\frac{2017}{2}$ + $\frac{2018}{1}$ = $(\frac{1}{2018} + 1)$ + $(\frac{2}{2017} + 1)$ + $(\frac{3}{2016} + 1)$ + … + $(\frac{2016}{3} + 1)$ + $(\frac{2017}{2} + 1)$ + $(\frac{2018}{1} + 1)$ = $\frac{2019}{2018}$ + $\frac{2019}{2017}$ + $\frac{2019}{2016}$ + … + $\frac{2019}{3}$ + $\frac{2019}{2}$ + $\frac{2019}{2019}$
=> $(Q-2019.P)^{2019}$ = [($\frac{2019}{2018}$ + $\frac{2019}{2017}$ + $\frac{2019}{2016}$ + … + $\frac{2019}{3}$ + $\frac{2019}{2}$ + $\frac{2019}{2019}$) – ($\frac{2019}{2}$ $+$ $\frac{2019}{3}$ $+$ $…$ $+$ $\frac{2019}{2019})]^{2019}$ = $0^{2}$ = $0$
Vậy $(Q-2019.P)^{2019}$ = $0$
$Q=\frac{1}{2018}+\frac{2}{2017}+…+\frac{2017}{2}+\frac{2018}{1}$
$=(\frac{1}{2018}+1)+(\frac{2}{2017}+1)+…+(\frac{2017}{2}+1)+1$
$=\frac{2019}{2018}+\frac{2019}{2017}+…+\frac{2019}{2}+\frac{2019}{2019}$
$=\frac{2019}{2019}+\frac{2019}{2018}+\frac{2019}{2017}+…+\frac{2019}{2}$
Ta có: $(Q-2019.P)^{2019}=[\frac{2019}{2019}+\frac{2019}{2018}+\frac{2019}{2017}+…+\frac{2019}{2}-2019.(\frac{1}{2}+\frac{1}{3}+…+\frac{1}{2018}+\frac{1}{2019})]^{2019}$
$=(\frac{2019}{2019}+\frac{2019}{2018}+\frac{2019}{2017}+…+\frac{2019}{2}-\frac{2019}{2019}-\frac{2019}{2018}-\frac{2019}{2017}-…-\frac{2019}{2})^{2019}$
$=0^{2019}=0$.