Question

The value of ${\sum }_{r=1}^n(-1{)}^{r-1}(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{r}){}^n{C}_r$ is equal to

• A. -1
• B. $\frac{1}{r}$
• C. $\frac{1}{n}$
• D. $-\frac{1}{n^3}$

Answer 1

The correct option is C

$\frac{1}{n}$

${\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{r})$

$={\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r\cdot {[x+\frac{x^2}{2}+\frac{x^3}{3}+....+\frac{x^r}{r}]}_0^1$

$={\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r{\int }_0^1(1+x+x^2+...+x^{r-1}dx)$

$={\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r{\int }_0^1\left(\frac{1-x^r}{1-x}\right)dx$

$={\int }_0^1{\sum }_{r=1}^n\frac{(-1{)}^{r-1}{\cdot }^{n{C}_r}-(-1{)}^{r-1}{\cdot }^{n{C}_r}{x}^r}{1-x}dx$

$={\int }_0^1\frac{{\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r-{\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r{x}^r}{1-x}dx$

Now, ${\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r={}^n{C}_1-{}^n{C}_2+{}^n{C}_3-{}^n{C}_4+\cdots ={}^n{C}_0-[{}^n{C}_0-{}^n{C}_1+{}^n{C}_2-{}^n{C}_3+{}^n{C}_4-\cdots ]=1-0=1$

Also, ${\sum }_{r=1}^n(-1{)}^{r-1}\cdot {}^n{C}_r{x}^r={}^n{C}_{1}x-{}^n{C}_2{x}^2+{}^n{C}_3{x}^3-\cdots +(-1{)}^{r-1}{}^n{C}_n{x}^n={}^n{C}_0-[{}^n{C}_0-{}^n{C}_{1}x+{}^n{C}_2{x}^2-{}^n{C}_3{x}^3+\cdots +(-1{)}^{r-1}{}^n{C}_n{x}^n]=1-(1-x{)}^n$

$\therefore {\sum }_{r=1}^n(-1{)}^{r-1}(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{r}){}^n{C}_r={\int }_0^1\frac{1-1+(1-x{)}^n}{1-x}dx={\int }_0^1(1-x{)}^{n-1}dx=\frac{1}{n}$