Question

c=3^0+3^1+3^2+3^2+...+3^42=?​

Answer 1

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Answer 2

 

$\displaystyle\bf\\C=3^0+3^1+3^2+3^3+...+3^{42}=?\\\\Folosim~urmatoarea~formula:\\\\k^0+k^1+k^2+k^3+...+k^n=\frac{k^{n+1}-1}{k-1}\\\\\textbf{Aceasta formula e valabila doar daca primul exponent este 0 (zero).}\\\textbf{Daca primul exponent este mai mare de 0, atunci folosim}\\\textbf{formula de 2 ori.}\\\\\\C=3^0+3^1+3^2+3^3+...+3^{42}=\frac{3^{42+1}-1}{3-1}=\boxed{\bf\frac{3^{43}-1}{2}}$