Binomial Theorem Question - None

Question

If C₀, C₁, C₂, ..., C_n denote the binomial coefficients in the expansion of (1 + x)ⁿ, then $\sum_\limits{r=0}^{n}$ (-1)^r ⁿC_r = $\frac{1+r \log _{e} 10}{\left(1+\log _{e} 10^{n}\right)^{r}}$ is equal to

Accepted Answer

Let log_e 10 = x. Then,
$\sum_\limits{r=0}^{n}$ (-1)^r ⁿC_r $\frac{1+r \log _{e} 10}{\left(1+\log _{e} 10^{n}\right)^{r}}$
= $\sum_\limits{r=0}^{n}$ (-1)^r ⁿC_r $\frac{1+r x}{(1+n x)^{r}}$
= $\sum_\limits{r=0}^{n}$ (-1)^r ⁿC_r $\left(\frac{1}{1+n x}\right)^{r}$ + $\sum_\limits{r=1}^{n}$ (-1)^r $\frac{n}{r}$ (^n-1C_r-1) $\frac{r x}{(1+n x)^{r}}$
= $\sum_\limits{r=0}^{n}$ (-1)^r ⁿC_r $\left(\frac{1}{1+n x}\right)^{r}$- $\frac{n x}{1+n x} \cdot \sum_\limits{r=1}^{n}$ (-1)^r-1 ^n-1C_r-1 $\left(\frac{1}{1+n x}\right)^{r-1}$
$=\left(1-\frac{1}{1+n x}\right)^{n}-\frac{n x}{1+n x}\left(1-\frac{1}{1+n x}\right)^{n-1}$
$=\left(\frac{n x}{1+n x}\right)^{n}-\left(\frac{n x}{1+n x}\right)^{n}$ = 0

Question:

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