MCQ
The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ is equal to
- ✓$\frac{{{2^n} - 1}}{{n + 1}}$
- B$n{.2^n}$
- C$\frac{{{2^n}}}{n}$
- D$\frac{{{2^n} + 1}}{{n + 1}}$
$\frac{{{{(1 + x)}^n} - {{(1 - x)}^n}}}{2} = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + ....$
Integrating from $x = 0$ to $x = 1$, we get$\frac{1}{2}\int\limits_0^1 {\{ {{(1 + x)}^n} - {{(1 - x)}^n}\} \,} dx$
$ = \int\limits_0^1 {({C_1}x + {C_3}{x^3} + {C_5}{x^5} + ....)} dx$
==> $\frac{1}{2}\left\{ {\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}} + \frac{{{{(1 - x)}^{n + 1}}}}{{n + 1}}} \right\}_0^1 = \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + ....$
or $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .... = \frac{1}{2}\left\{ {\frac{{{2^{n + 1}} - 1}}{{n + 1}} + \frac{{0 - 1}}{{n + 1}}} \right\}$
$ = \frac{1}{2}\left( {\frac{{{2^{n + 1}} - 2}}{{n + 1}}} \right) = \frac{{{2^n} - 1}}{{n + 1}}$
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$(i)$ Reflection about the line $y = x$
$(ii)$ Translation through a distance $2$ units along the positive $x$-axis
Then the final coordinates of the point are