Question Number 56416 by gunawan last updated on 16/Mar/19
$$\underset{{n}−\mathrm{digits}} {\left(\mathrm{666}\:….\:\mathrm{6}\right)^{\mathrm{2}} }\:+\:\underset{{n}−\mathrm{digits}} {\left(\mathrm{888}\:….\mathrm{8}\right)}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by mr W last updated on 16/Mar/19
$$=\underset{\mathrm{2}{n}\:{digits}} {\left(\mathrm{444}…\mathrm{4}\right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
![(666....6)←n digit 6(111...1) 6{10^(n−1) +10^(n−2) +10^(n−3) +...+1} 6×((10^(n−1) (1−(1/(10^n ))))/((1−(1/(10)))))→(6/9)×10^n (((10^n −1)/(10^n )))→(6/9)(10^n −1) so answer is ((36)/(81))(10^n −1)^2 +(8/9)(10^n −1) {(4/9)(10^n −1)^2 }+(8/9)(10^n −1) (4/9)(10^n −1)(10^n −1+2) (4/9)(10^(2n) −1) 4(((10^(2n) −1)/(10−1))) [a(((r^n −1)/(r−1)))=a+ar+ar^2 +...+ar^n ] 4(1+10+10^2 +10^3 +...+10^(2n) ) (4+40+400+4000+.....) (444....4)←2n digit](https://www.tinkutara.com/question/Q56444.png)
$$\left(\mathrm{666}….\mathrm{6}\right)\leftarrow{n}\:{digit} \\ $$$$\mathrm{6}\left(\mathrm{111}…\mathrm{1}\right) \\ $$$$\mathrm{6}\left\{\mathrm{10}^{{n}−\mathrm{1}} +\mathrm{10}^{{n}−\mathrm{2}} +\mathrm{10}^{{n}−\mathrm{3}} +…+\mathrm{1}\right\} \\ $$$$\mathrm{6}×\frac{\mathrm{10}^{{n}−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}^{{n}} }\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right)}\rightarrow\frac{\mathrm{6}}{\mathrm{9}}×\mathrm{10}^{{n}} \left(\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{10}^{{n}} }\right)\rightarrow\frac{\mathrm{6}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$${so}\:{answer}\:{is} \\ $$$$\frac{\mathrm{36}}{\mathrm{81}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\left\{\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} \right\}+\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)\left(\mathrm{10}^{{n}} −\mathrm{1}+\mathrm{2}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right) \\ $$$$\mathrm{4}\left(\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{10}−\mathrm{1}}\right) \\ $$$$\left[{a}\left(\frac{{r}^{{n}} −\mathrm{1}}{{r}−\mathrm{1}}\right)={a}+{ar}+{ar}^{\mathrm{2}} +…+{ar}^{{n}} \right] \\ $$$$\mathrm{4}\left(\mathrm{1}+\mathrm{10}+\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} +…+\mathrm{10}^{\mathrm{2}{n}} \right) \\ $$$$\left(\mathrm{4}+\mathrm{40}+\mathrm{400}+\mathrm{4000}+…..\right) \\ $$$$\left(\mathrm{444}….\mathrm{4}\right)\leftarrow\mathrm{2}{n}\:{digit} \\ $$
Commented by mr W last updated on 16/Mar/19
$${good}\:{job}\:{sir}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
$${thsnk}\:{you}\:{sir}… \\ $$