Question

Prove that $\frac{C_1}{1}-\frac{C_2}{2}+\frac{C_3}{3}-\frac{C_4}{4}+.....+\frac{(-1{)}^{n-1}}{n}{C}_n$
$=1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}.$

Answer 1

$(1-x{)}^n=1-C_{1}x+C_2{x}^2-C_3{x}^3+(-1{)}^n{C}_n{x}^n.$
or $\frac{(1-x{)}^n}{x}-\frac{1}{x}=-C_1+C_{2}x-C_3{x}^2+...+(-1{)}^n{C}_n{x}^{n-1}$
$\frac{1-(1-x{)}^n}{1-(1-x)}=-C_1+C_{2}x-C_3{x}^2+...+(-1{)}^n{C}_n{x}^{n-1}$
L.H.S. = Sum of a G.P.
$\therefore 1+(1-x)+(1-x{)}^2+......+(1-x{)}^{n-1}$
$C_1-C_{2}x+C_3{x}^2...........+(-1{)}^{n-1}{C}_n{x}^{n-1}$
Putting x = 0, we get
$-\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{n}+C=0$
$\therefore C=\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}$
Now putting x = 1 in both sides of (1)
$1+C=C_1-\frac{C_2}{2}+\frac{c_3}{3}-\frac{C_4}{4}+.......$
Now put for C from (2)
or $1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}$
$=C_1-\frac{C_2}{2}+\frac{C_3}{3}-+(-1{)}^{n-1}\frac{C_n}{n}$