Question

Fie S=1+2*5+3*5^{2}+4*5^{3}+...+200*5^{199} :
a) Calculeaza S;
b) Demonstrati ca numarul (799*5^{200}+1) este divizibil cu 16.

Answer 1

[tex]S(x)=x+x^2+x^3+...+x^n=x* \frac{x^{n}-1}{x-1}= \frac{x^{n+1}-x}{x-1}\\ S'(x)=1+2x+3x^2+...+nx^{n-1}=( \frac{x^{n+1}-x}{x-1})'=\\ =\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} \\ Pentru\ n=200\ si\ x=5=\ \textgreater \ \\ [/tex]
[tex]S=1+2*5+3*5^{2}+4*5^{3}+...+200*5^{199}=S'(5)=\\ = \frac{200*5^{201}-201*5^{200}+1}{16}=\frac{5^{200}(1000-201)+1}{16}=\\ = \frac{799*5^{200}+1}{16}\\ Deoarece S\in Z =\ \textgreater \ (799*5^{200}+1)\vdots16\\ [/tex]