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Sum of Coefficients of All Terms

If C0, C1 ,C2...

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ denote the binomial coefficients in the expansion of $(1 + x)^{n}$ and $k = n \sum r = 0 (- 1)^{r}^{n} C_{r} \frac{1 + r {log}_{e} 10}{(1 + {log}_{e} 10^{n})^{r}},$ then $| k + 3 |$ is equal to

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Solution

Let ${log}_{e} 10 = x .$ Then,
$n \sum r = 0 (- 1)^{r}^{n} C_{r} \frac{1 + r {log}_{e} 10}{(1 + {log}_{e} 10^{n})^{r}}$
$= n \sum r = 0 (- 1)^{r}^{n} C_{r} \frac{1 + r x}{(1 + n x)^{r}}$
$= n \sum r = 0 (- 1)^{r}^{n} C_{r} {(\frac{1}{n x + 1})}^{r} + n \sum r = 0 (- 1)^{r} \frac{n}{r}^{n - 1} C_{r - 1} \frac{r x}{(n x + 1)^{r}}$
$= n \sum r = 0 (- 1)^{r}^{n} C_{r} {(\frac{1}{n x + 1})}^{r} - \frac{n x}{1 + n x} \cdot n \sum r = 1 (- 1)^{r - 1}^{n - 1} C_{r - 1} \frac{1}{(n x + 1)^{r - 1}}$
$= {(1 - \frac{1}{1 + n x})}^{n} - \frac{n x}{1 + n x} {(1 - \frac{1}{1 + n x})}^{n - 1}$
$= {(\frac{n x}{1 + n x})}^{n} - {(\frac{n x}{1 + n x})}^{n} = 0$

$∴ | k + 3 | = 3$

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Sum of Coefficients of All Terms

Standard XII Mathematics

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