Question

cum se calculeaza suma 1+2x5+3x5²+4x5³+....+200x5la puterea199?

Answer 1

$\textbf{Metoda 1:}$

$S =1+2\cdot 5+3\cdot 5^2+4\cdot 5^3+...+200\cdot 5^{199}$

$S(x) =\displaystyle \sum\limits_{k=1}^{200} k\cdot x^{k-1} \Big|\int \Rightarrow$

$\displaystyle \begin{aligned}\Rightarrow \int S(x)\,dx &= \int \sum\limits_{k=1}^{200}k\cdot x^{k-1}\, dx \\ &=\sum\limits_{k=1}^{200}k\cdot \dfrac{x^{k-1+1}}{k-1+1} +C \\& =\sum\limits_{k=1}^{200} x^k + C\\ &= x^1+x^2+x^3+...+x^{200} +C\\ &= x\cdot \dfrac{x^{200}-1}{x-1} +C\\ &= \dfrac{x^{201}-x}{x-1}+C\end{aligned}$

$\displaystyle \left(\int S(x)\, dx\right)' =\left(\dfrac{x^{201}-x}{x-1}+C\right)'$

$\Rightarrow S(x) = \dfrac{(201x^{200}-1)(x-1)-(x^{201}-1)}{(x-1)^2}$

$\Rightarrow S(5) = \dfrac{201\cdot 5^{200}\cdot (5-1)-(5^{201}-1)}{(5-1)^2}$

$\Rightarrow S(5) =\dfrac{804\cdot 5^{200}-5^{201}+1}{16}$

$\Rightarrow \boxed{S = \dfrac{799\cdot 5^{200}+1}{16}}$

$\textbf{Metoda 2:}$

$S =1+2\cdot 5+3\cdot 5^2+4\cdot 5^3+...+200\cdot 5^{199}\\ 5S =\,\,\,\,\,\,\,\,\,\,\,\,\,\, 5+2\cdot 5^2+3\cdot 5^3+...+199\cdot 5^{199}+200\cdot 5^{200}$

$S-5S = 5\cdot (2-1)+5^2\cdot (3-2)+5^3\cdot (4-3)+...+5^{199}\cdot (200-199)-\\ -200\cdot 5^{200}+1$

$-4S = 5+5^2+5^3+...+5^{199}-200\cdot 5^{200}+1$

$-4S = 5\cdot \dfrac{5^{199}-1}{5-1}-200\cdot 5^{200}+1$

$-4S = \dfrac{5^{200} - 5}{4}-200\cdot 5^{200}+1$

$-4S = \dfrac{5^{200}-5-800\cdot 5^{200}+4}{4}$

$-S = \dfrac{-799\cdot 5^{200}-1}{16}$

$S = \dfrac{799\cdot 5^{200}+1}{16}$

Answer 2

Am atașat o rezolvare.