Question

b Calculați: 9 + 99 +999 +...+999...9 +2012=
2011 cifre​

Answer 1

Progresie geometrica:

q=ratie

n=numar termeni

b₁, b₂,..., bₙ - termenii

Suma unei progresii geometrice:

$S_n=b_1\cdot \frac{q^n-1}{q-1}$

Observam ca:

$9=10-1\\\\99=10^2-1\\\\999=10^3-1\\\\...\\\\999...999=10^{2011}-1$

Le adunam si obtinem:

9+99+999+....+999...9999 =10-1+10²-1+10³-1+...+10²⁰¹¹-1=-2011+10+10²+10³+...+10²⁰¹¹

Luam separat: 10+10²+10³+...+10²⁰¹¹

Avem suma unei progresii geometrice

b₁=10

n=2011

q=10

$S_n=10\cdot \frac{10^{2011}-1}{10-1} =\frac{10^{2012}-10}{9}$

$9+99+999+...+999...999+2012=-2011+\frac{10^{2012}-10}{9} +2012=1+\frac{10^{2012}-10}{9}$

Un alt exercitiu gasesti aici: https://brainly.ro/tema/1883996

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