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Question Number 56416 by gunawan last updated on 16/Mar/19

(666 .... 6)^2 _(n−digits)  + (888 ....8)_(n−digits)  is equal to

$$\underset{{n}−\mathrm{digits}} {\left(\mathrm{666}\:....\:\mathrm{6}\right)^{\mathrm{2}} }\:+\:\underset{{n}−\mathrm{digits}} {\left(\mathrm{888}\:....\mathrm{8}\right)}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Commented by mr W last updated on 16/Mar/19

=(444...4)_(2n digits)

$$=\underset{\mathrm{2}{n}\:{digits}} {\left(\mathrm{444}...\mathrm{4}\right)} \\ $$

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

$$\left(\mathrm{666}....\mathrm{6}\right)\leftarrow{n}\:{digit} \\ $$$$\mathrm{6}\left(\mathrm{111}...\mathrm{1}\right) \\ $$$$\mathrm{6}\left\{\mathrm{10}^{{n}−\mathrm{1}} +\mathrm{10}^{{n}−\mathrm{2}} +\mathrm{10}^{{n}−\mathrm{3}} +...+\mathrm{1}\right\} \\ $$$$\mathrm{6}×\frac{\mathrm{10}^{{n}−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}^{{n}} }\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right)}\rightarrow\frac{\mathrm{6}}{\mathrm{9}}×\mathrm{10}^{{n}} \left(\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{10}^{{n}} }\right)\rightarrow\frac{\mathrm{6}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$${so}\:{answer}\:{is} \\ $$$$\frac{\mathrm{36}}{\mathrm{81}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\left\{\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} \right\}+\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)\left(\mathrm{10}^{{n}} −\mathrm{1}+\mathrm{2}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right) \\ $$$$\mathrm{4}\left(\frac{\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{10}−\mathrm{1}}\right) \\ $$$$\left[{a}\left(\frac{{r}^{{n}} −\mathrm{1}}{{r}−\mathrm{1}}\right)={a}+{ar}+{ar}^{\mathrm{2}} +...+{ar}^{{n}} \right] \\ $$$$\mathrm{4}\left(\mathrm{1}+\mathrm{10}+\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} +...+\mathrm{10}^{\mathrm{2}{n}} \right) \\ $$$$\left(\mathrm{4}+\mathrm{40}+\mathrm{400}+\mathrm{4000}+.....\right) \\ $$$$\left(\mathrm{444}....\mathrm{4}\right)\leftarrow\mathrm{2}{n}\:{digit} \\ $$

Commented by mr W last updated on 16/Mar/19

good job sir!

$${good}\:{job}\:{sir}! \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

thsnk you sir...

$${thsnk}\:{you}\:{sir}... \\ $$

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