Question

3 la puterea a 2 + 3 la puterea a 2a .... 3 la puterea 100

Answer 1

Notam suma cu S.

$S=3+3^2+\dots+3^{100}|\cdot 3\\3S=3^2+3^3+\ldots+ 3^{101}\\-----------\\3S-S=3^{101}-3\\2S=3^{101}-3\\\boxed{S=\dfrac{3^{101}-3}{2}}$

Answer 2

S=3+3²+3³+...+3¹⁰⁰ ,observam ca termenii sumei reprezinta o progresie geometrica de ratie q=(3²/3)=(3³/3²)=...=(3¹⁰⁰/3⁹⁹)=3 =>

S=b₁·(qⁿ -1) /q-1 ,unde n reprezinta numarul de termenii din progresie (in cazul nostru n=100) => S=3·(3¹⁰⁰ -1) /3-1 => S=3·(3¹⁰⁰ -1) /2 .