Question

Calculati : 9+99+999+....+999_9(2011 cifre) +2012
Multumesc!!

Answer 1

$\displaystyle 9+99+...+\underbrace{999...9}_{\mbox{2011}}+2012= \\ \\ =(9+1)+(99+1)+...+(\underbrace{999...9}_{\mbox{2011}}+1)+1= \\ \\ =10+10^2+...+10^{2011}+1= \\ \\ =1+10+10^2+...+10^{2011}= \\ \\ =\frac{10^{2012}-1}{9}.$

Answer 2

S = 9+99+999+....+999...9 + 2012 = 10 - 1 + 100 - 1 + 1000 - 1 + ... + 10...0 - 1 + 2012.

10...0 are 2011 zerouri, iar - 1 apare de 2011 ori. Deci:

S = 10 + 10² + 10³ + ... + 10²⁰¹¹ - 2011 + 2012 = 10 + 10² + 10³ + ... + 10²⁰¹¹ + 1.

Notăm cu S₁ = 10 + 10² + 10³ + ... + 10²⁰¹¹. Avem că:

10S₁ = 10² + 10³ + 10⁴ + ... + 10²⁰¹¹ + 10²⁰¹².

10S₁ - S₁ = 9S₁ = (10² + 10³ + 10⁴ + ... + 10²⁰¹¹ + 10²⁰¹²) - (10 + 10² + 10³ + ... + 10²⁰¹¹) = 10²⁰¹² - 10, deci S₁ = (10²⁰¹² - 10) / 9

La final:

S = S₁ + 1 = (10²⁰¹² - 10) / 9 + 1 = (10²⁰¹² - 10) / 9 + 9 / 9 = (10²⁰¹² - 1) / 9.

S = (10²⁰¹² - 1) / 9.

Simplu, nu ? :-))).

Green eyes.