Question

S=1+3+3 la puterea 2+...+3 la puterea 100. Atentie! Suma lui Gauss.

Answer 1

$S = 1+3+3^2+...+3^{99}+3^{100} \\ S = 3^0+3^1+3^2+...+3^{99}+3^{100} \Big|\cdot 3 \\ 3\cdot S = 3\cdot( 3^0+3^1+3^2+...+3^{99}+3^{100}) \\ 3\cdot S = 3^1+3^2+3^3+...+3^{100}+3^{101} \\ 3\cdot S = 3^0-3^0+3^1+3^2+3^3+...+3^{100}+3^{101} \\ 3\cdot S = 3^0+3^1+3^2+3^3+...+3^{100}+3^{101} -3^0 \\ \\ (Observam ca 3^0+3^1+3^2+3^3+...+3^{100} este chiar S, inlocuim acea suma cu S) \\ \\ \Rightarrow 3\cdot S = S + 3^{101}-3^0 \Rightarrow 3\cdot S - S = 3^{101}-1 \Rightarrow 2\cdot S = 3^{101}-1 \Rightarrow$

$\Rightarrow \boxed{\boxed{S =\dfrac{3^{101}-1}{2}} }$