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Question Number 6831 by Tawakalitu. last updated on 30/Jul/16
Find the 100th digits of this sum 1 + 50 + 50^2 + ..... + 50^(999)
$${Find}\:{the}\:\mathrm{100}{th}\:{digits}\:{of}\:{this}\:{sum}\: \\ $$$$\mathrm{1}\:+\:\mathrm{50}\:+\:\mathrm{50}^{\mathrm{2}} \:+\:…..\:+\:\mathrm{50}^{\mathrm{999}} \\ $$
Commented by prakash jain last updated on 31/Jul/16
How did you get 5^(1000) ×10^(1000) −1=(5^(1000) −1)×(999...9999)?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}=\left(\mathrm{5}^{\mathrm{1000}} −\mathrm{1}\right)×\left(\mathrm{999}…\mathrm{9999}\right)? \\ $$
Commented by Yozzii last updated on 30/Jul/16
Σ_(i=0) ^(999) 50^i =((50^(1000) −1)/(50−1))=((50^(1000) −1)/(49)) =((5^(1000) ×10^(1000) −1)/(49)) Now, n×10^m =n000...000 (m zeroes) (n∈N) ⇒n10^m −1=(n−1)999...999 (m nines) ∴ 5^(1000) ×10^(1000) =(5^(1000) )000...000 (1000 zeroes trailing) ⇒5^(1000) ×10^(1000) −1=(5^(1000) −1)999...999 (1000 nines trailing) (continue)
$$\underset{{i}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\mathrm{50}^{{i}} =\frac{\mathrm{50}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{50}−\mathrm{1}}=\frac{\mathrm{50}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{49}} \\ $$$$=\frac{\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}}{\mathrm{49}} \\ $$$${Now},\:{n}×\mathrm{10}^{{m}} ={n}\mathrm{000}…\mathrm{000}\:\:\left({m}\:{zeroes}\right)\:\left({n}\in\mathbb{N}\right) \\ $$$$\Rightarrow{n}\mathrm{10}^{{m}} −\mathrm{1}=\left({n}−\mathrm{1}\right)\mathrm{999}…\mathrm{999}\:\left({m}\:{nines}\right) \\ $$$$ \\ $$$$\therefore\:\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} =\left(\mathrm{5}^{\mathrm{1000}} \right)\mathrm{000}…\mathrm{000}\:\left(\mathrm{1000}\:{zeroes}\:{trailing}\right) \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{1000}} ×\mathrm{10}^{\mathrm{1000}} −\mathrm{1}=\left(\mathrm{5}^{\mathrm{1000}} −\mathrm{1}\right)\mathrm{999}…\mathrm{999}\:\left(\mathrm{1000}\:{nines}\:{trailing}\right) \\ $$$$\left({continue}\right) \\ $$
Commented by Yozzii last updated on 31/Jul/16
I was trying to write the number in decimal form. e.g 5^2 ×10^3 =25000 ⇒5^2 ×10^3 −1=24999=(5^2 −1)999 (in decimal representation, not a product)
$${I}\:{was}\:{trying}\:{to}\:{write}\:{the}\:{number} \\ $$$${in}\:{decimal}\:{form}. \\ $$$${e}.{g}\:\mathrm{5}^{\mathrm{2}} ×\mathrm{10}^{\mathrm{3}} =\mathrm{25000} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2}} ×\mathrm{10}^{\mathrm{3}} −\mathrm{1}=\mathrm{24999}=\left(\mathrm{5}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{999}\:\:\left({in}\:{decimal}\right. \\ $$$$\left.{representation},\:{not}\:{a}\:{product}\right) \\ $$
Commented by Tawakalitu. last updated on 31/Jul/16
Thanks sir
$${Thanks}\:{sir} \\ $$
Commented by Tawakalitu. last updated on 31/Jul/16
Thanks sir
$${Thanks}\:{sir} \\ $$

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