+0

0
161
1

Compute

$$1 + \frac{1}{3} + \frac{2}{3} + \frac{2}{9} + \frac{4}{9} + \frac{4}{27} + \frac{8}{27} + \frac{8}{81} + \dotsb$$

compute $$1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007$$

Guest Feb 26, 2017
edited by Guest  Feb 26, 2017
#1
+19653
0

compute

1 - 2 + 3 - 4 + $$\dots$$+ 2005 - 2006 + 2007 = ?

Geometric series:

$$\begin{array}{|rcll|} \hline 1+x+x^2+x^3+x^4+x^5+\dots + x^n &=& \dfrac{1-x^{n+1}}{1-x} \qquad -1 < x < 1 \\ \hline \end{array}$$

Derivative:

$$\begin{array}{|rcll|} \hline 1+2x+3x^2+4x^3+5x^4+6x^5+\dots + nx^{n-1} &=& \dfrac{1-x^n[~n(1-x)+1~]}{(1-x)^2} \\ \hline \end{array}$$

x $$\rightarrow$$ -x

$$\begin{array}{|rcll|} \hline 1-2x+3x^2-4x^3+5x^4-6x^5+-\dots + n(-x)^{n-1} &=& \dfrac{1-(-x)^n[~n(1+x)+1~]}{(1+x)^2} \\ \hline \end{array}$$

x = 1:

$$\begin{array}{|rcll|} \hline 1-2+3-4+5-6+-\dots + n(-1)^{n-1} &=& \dfrac{1-(-1)^n[~2n+1~]}{4} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^{2007}n (-1)^{n - 1} = 1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007 \\\\ &=& \dfrac{1-(-1)^{2007}[~2\cdot2007+1~]}{4} \\\\ &=& \dfrac{1+4015}{4} \\\\ &=& \dfrac{4016}{4} \\\\ &\mathbf{=}& \mathbf{1004} \\ \hline \end{array}$$

heureka  Feb 27, 2017