Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Required value is $\frac{{10}_{c_1}}{1}-\frac{{10}_{c_2}}{2}+\frac{{10}_{c_3}}{3}-----\frac{{10}_{c_{10}}}{10}$
To find which, consider $(1-x{)}^{10}={10}_{c_0}-{10}_{c_1}x+{10}_{c_2}{x}^2------+{10}_{c_{10}}{x}^{10}\Rightarrow \frac{(1-x{)}^{10}-1}{x}=-[{10}_{c_1}-{10}_{c_2}x+----+x_{c_{10}}{x}^9]\Rightarrow {\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx={\int }_0^1[{10}_{c_1}-{10}_{c_2}x+-----+{10}_{c_{10}}{x}^9]dx$
$={10}_{c_1}\frac{{10}_{c_2}}{2}+\frac{{10}_{c_3}}{3}+---------\frac{10c_{10}}{10}$
To find LHS consider $I_n={\int }_0^1\frac{1-(1-x{)}^n}{x}dx\Rightarrow I_{n+1}-I_n={\int }_0^1(1-x{)}^{n}dx=\frac{1}{n+1}$
$\therefore I_{n+1}=\frac{1}{n+1}+I_n\therefore I_{10}={\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx=\frac{1}{10}+I_9=\frac{1}{10}+\frac{1}{9}+I_8\approx ----=\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+-----+1$