Question

If $n\in N$, then ${\sum }_{r=0}^n{(-1)}^r.{{}^{n}C}_r\frac{(1+{\log }_{e}10)}{{(1+{{\log }_{e}10}^n)}^r}=$

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $0$

Let ${\log }_{e}10=x$
Then $\sum _{r=0}^n{(-1)}^r{.}^n{C}_r.\frac{1+r{\log }_{e}10}{{(1+{\log }_{e}10)}^r}=\sum _{r=0}^n{(-1)}^r{.}^n{C}_r.\frac{1+rx}{{(1+nx)}^r}$
$=\sum _{r=0}^n{(-1)}^r{.}^n{C}_r{\left(\frac{1}{1+nx}\right)}^r+\sum _{r=0}^n{(-1)}^r\frac{n}{r}{.}^{n-1}{C}_{r-1}\frac{rx}{{(1+nx)}^r}$
${=}^n{C}_r{\left(\frac{1}{1+nx}\right)}^r-\frac{nx}{1+nx}\sum _{r=0}^n{(-1)}^{r-1}{{.}^{n-1}C}_{r-1}{\left(\frac{1}{1+nx}\right)}^{r-1}$
$={(1-\frac{1}{1+nx})}^n-\frac{nx}{1+nx}{(1-\frac{1}{1+nx})}^{n-1}$
$={\left(\frac{nx}{1+nx}\right)}^n-{\left(\frac{nx}{1+nx}\right)}^n=0$