Question

If $C_0,C_1,C_2,.....C_r$ are binomial coefficients in the expansion of ${(1+x)}^n$ then
$C_1-\frac{C_2}{2}+\frac{C_3}{3}-\frac{C_4}{4}+....{(-1)}^{n-1}\frac{C_n}{n}$ equals

• A. ${\sum }_{r=1}^n{(-1)}^{n-1}\frac{C_r}{r}$
• B. ${\sum }_{r=1}^n\frac{1}{r}$
• C. ${\sum }_{r=2}^n\frac{1}{r-1}$
• D. None of these

Answer 1

Answer: (B)

${\sum }_{r=1}^n\frac{1}{r}$

Consider ${(1-x)}^n=C_0-C_{1}x+C_2{x}^2-C_3{x}^3+......$

$\Rightarrow 1-{(1-x)}^n=C_{1}x-C_2{x}^2+C_3{x}^3-......$

integrating both side w.r.t $x$ with limit $0$ to $1$

$\therefore {\int }_0^1(C_{1}x-C_2{x}^2+C_3{x}^3-.....)dx={\int }_0^1\frac{1-{(1-x)}^n}{1-(1-x)}dx$

$\Rightarrow \frac{C_1}{1}-\frac{C_2}{2}+\frac{C_3}{3}+....{(-1)}^n-1\frac{C_n}{n}={\int }_0^1\frac{1-x^n}{1-x}dx$

$(Using{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$

$={\int }_0^1(1+x+x^2+...+x^{n-1})dx$

$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{n}={\sum }_{r=1}^n\frac{1}{r}$