Mathematics - Binomial Theorem Question with Solution | TestHub

The required value is $C_{10}{C}_1-\frac{C_{10}{C}_2}{2}+\frac{C_{10}{C}_3}{3}-\cdots -\frac{C_{10}{C}_{10}}{10}$.

To find this, consider the binomial expansion of $(1-x{)}^{10}$:

$$(1-x{)}^{10}=C_{10}{C}_0-C_{10}{C}_{1}x+C_{10}{C}_2{x}^2-\cdots +C_{10}{C}_{10}{x}^{10}$$

Rearranging the equation:

$$\frac{(1-x{)}^{10}-1}{x}=-\left[C_{10}{C}_1-C_{10}{C}_{2}x+\cdots +C_{10}{C}_{10}{x}^9\right]$$

Now, integrate both sides from $0$ to $1$:

$${\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx={\int }_0^1\left[C_{10}{C}_1-C_{10}{C}_{2}x+\cdots +C_{10}{C}_{10}{x}^9\right]dx$$

$$=C_{10}{C}_1-\frac{C_{10}{C}_2}{2}+\frac{C_{10}{C}_3}{3}-\cdots -\frac{C_{10}{C}_{10}}{10}$$

To find the Left Hand Side (LHS), consider the integral $I_n={\int }_0^1\frac{1-(1-x{)}^n}{x}dx$.

We can establish a recurrence relation:

$$\begin{gathered} \text{displaylines}{I}_{n+1}-I_n={\int }_0^1\frac{1-(1-x{)}^{n+1}-(1-(1-x{)}^n)}{x}dx= \\ {\int }_0^1\frac{(1-x{)}^n-(1-x{)}^{n+1}}{x}dx \end{gathered}$$

$$={\int }_0^1\frac{(1-x{)}^n(1-(1-x))}{x}dx={\int }_0^1\frac{(1-x{)}^{n}x}{x}dx={\int }_0^1(1-x{)}^{n}dx$$

$$={\left[-\frac{(1-x{)}^{n+1}}{n+1}\right]}_0^1=0-\left(-\frac{1}{n+1}\right)=\frac{1}{n+1}$$

Therefore, $I_{n+1}=I_n+\frac{1}{n+1}$.

Using this recurrence relation for $I_{10}$:

$$I_{10}={\int }_0^1\frac{1-(1-x{)}^{10}}{x}dx=I_9+\frac{1}{10}$$

$$I_9=I_8+\frac{1}{9}$$

Continuing this pattern:

$$I_{10}=\frac{1}{10}+\frac{1}{9}+I_8=\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+I_7=\dots$$

$$I_{10}=\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\cdots +\frac{1}{1}+I_0$$

Since $I_0={\int }_0^1\frac{1-(1-x{)}^0}{x}dx={\int }_0^1\frac{1-1}{x}dx=0$, we have:

$$I_{10}=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{10}$$