Question

${\sum }_{k=1}^{10}\frac{(-1{)}^{k-1}}{k}.\left({10}_{C_k}\right)=$

• A. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+-----+\frac{1}{11}$
  
• B. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+----+\frac{1}{10}$
  
• C. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+----+\frac{1}{9}$
  
• D. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+----+\frac{1}{12}$

Answer 1

The correct option is B $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+----+\frac{1}{10}$


Required value is $\frac{{}^{10}{C}_1}{1}-\frac{{}^{10}{C}_2}{2}+\frac{{}^{10}{C}_3}{3}\dots \dots \frac{{}^{10}{C}_{10}}{10}$
To find which, consider $(1-x{)}^{10}{=}^{10}{C}_0{-}^{10}{C}_{1}x{+}^{10}{C}_2{x}^2\dots {\dots }^{10}{C}_{10}{x}^{10}$
$\Rightarrow \frac{(1-x{)}^{10}-1}{z}=-{[}^{10}{C}_1{-}^{10}{C}_{2}x+\dots \dots {+}^{10}{C}_{10}{x}^9]$
$\Rightarrow {\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx={\int }_0^1{[}^{10}{C}_1{-}^{10}{C}_{2}x+\dots \dots {+}^{10}{C}_{10}{x}^9]dx$
${=}^{10}{C}_1-\frac{{}^{10}{C}_2}{2}+\frac{{}^{10}{C}_3}{3}\dots \dots \frac{{}^{10}{C}_{10}}{10}$
To find LHS consider $I_n={\int }_0^1\frac{1-(1-x{)}^n}{x}dx\Rightarrow I_{n+1}-I_n={\int }_0^1(1-x{)}^{n}dx=\frac{1}{n+1}$
$\therefore I_{n+1}=\frac{1}{n+1}+I_n$
$\therefore I_{10}={\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx=\frac{1}{10}+I_9=\frac{1}{10}+\frac{1}{9}+I_8\approx ----=\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+-----+1$