Question

Let
$S=C_1-(1+\frac{1}{2})C_2+(1+\frac{1}{2}+\frac{1}{3})C_3-........+.{(-1)}^{n-1}(1+\frac{1}{2}+....+\frac{1}{n})C_n$
then

• A. $nS=1$
• B. $\frac{1}{S}$ is an integer
• C. $\frac{1}{S^2}$ is an integer
• D. $S$ is independent of $n$

Answer 1

Answer: (A), (B), (C)

$\frac{1}{S^2}$ is an integer
$nS=1$
$\frac{1}{S}$ is an integer

We can write $S$ as
$S=a_1+\frac{1}{2}{a}_2+\frac{1}{3}{a}_3+.....+\frac{1}{n}{a}_n$
where
$a_r={(-1)}^{r-1}(C_r-C_{r-1}+.....+.{(-1)}^{n-r}{C}_n)={(-1)}^{n-1}(C_0-C_1+.....+.{(-1)}^{n-r}{C}_{n-r})[use{C}_r=C_{n-r}]$
But for $k\ge 0$
$C_0-C_1+C_2-.....+{(-1)}^k{C}_k$
$=$ Coefficient of $x^k$ in
$(C_0-C_{1}x+C_2{x}^2-.....+{(-1)}^n{C}_n{x}^n)(1+x+x^2+....)$
$=$Coefficient of $x^k$ in ${(1-x)}^n{(1-x)}^{-1}$
$=$ coefficient of $x^k$ in ${(1-x)}^{n-1}$
$={(-1)}^k\left({{}^{n-1}C}_k\right)$
Thus, $a_r={(-1)}^{n-1}{(-1)}^{n-r}\left({{}^{n-1}C}_{n-r}\right)$
$={(-1)}^{r-1}\left({{}^{n-1}C}_{n-r}\right)$
Now
$S={{}^{n-1}C}_0-\frac{1}{2}\left({{}^{n-1}C}_1\right)+\frac{1}{3}\left({{}^{n-1}C}_2\right)-.....+\frac{{(-1)}^{n-1}}{n}\left({{}^{n-1}C}_{n-1}\right)$
$={\int }_0^1{(1-x)}^{n-1}dx=\frac{1}{n}$