Question

If $C_0,C_1,C_2,\dots ,C_n$ denote the binomial coefficients respectively in $(1+x{)}^{2020},$ then

• A. $C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\dots +\frac{C_{2020}}{2021}=\frac{1}{2021}$
• B. $C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\dots +\frac{C_{2020}}{2021}=\frac{2^{2020}-1}{2021}$
• C. $C_0+\frac{C_2}{3}+\frac{C_4}{5}+\dots +\frac{C_{2020}}{2021}=\frac{2^{2020}}{2021}$
• D. $\frac{C_1}{2}+\frac{C_3}{4}+\dots +\frac{C_{1999}}{2020}=\frac{2^{2020}}{2021}$

Answer 1

Answer: (A), (C)

The correct options are
A $C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\dots +\frac{C_{2020}}{2021}=\frac{1}{2021}$
C $C_0+\frac{C_2}{3}+\frac{C_4}{5}+\dots +\frac{C_{2020}}{2021}=\frac{2^{2020}}{2021}$

Here $C_r={}^{2020}{C}_r$ and $n=2020$

$(1+x{)}^{2020}=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots +C_{2020}{x}^{2020}$

Integrating from $0$ to $1$ both sides,
$\Rightarrow \underset{0}{\overset{1}{\int }}(1+x{)}^{2020}dx=\underset{0}{\overset{1}{\int }}(C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots +C_{2020}{x}^{2020})dx$

$\Rightarrow {\left[\frac{(1+x{)}^{2021}}{2021}\right]}_0^1={[C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+\cdots +\frac{C_{2020}{x}^{2021}}{2021}]}_0^1$

$\Rightarrow \frac{2^{2021}}{2021}-\frac{1}{2021}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots +\frac{C_{2020}}{2021}$

$\therefore C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots +\frac{C_{2020}}{2021}=\frac{2^{2021}-1}{2021}\dots \left(i\right)$

Similarly,
$\underset{-1}{\overset{0}{\int }}(1+x{)}^{2020}dx=\underset{-1}{\overset{0}{\int }}(C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots +C_{2020}{x}^{2020})dx$
$\Rightarrow {\left[\frac{(1+x{)}^{2021}}{2021}\right]}_{-1}^0={[C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+\cdots +\frac{C_{2020}{x}^{2021}}{2021}]}_{-1}^0$
$\Rightarrow \frac{1}{2021}-0=0-(-C_0+\frac{C_1}{2}-\frac{C_2}{3}+\cdots -\frac{C_{2020}}{2021})$
$\therefore C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\cdots +\frac{C_{2020}}{2021}=\frac{1}{2021}\dots \left(ii\right)$

$\left(i\right)+\left(ii\right)$
$\Rightarrow 2(C_0+\frac{C_2}{3}+\frac{C_4}{5}+\cdots +\frac{C_{2020}}{2021})=\frac{2^{2021}}{2021}$
$\Rightarrow C_0+\frac{C_2}{3}+\frac{C_4}{5}+\cdots +\frac{C_{2020}}{2021}=\frac{2^{2020}}{2021}$

Similarly $\left(i\right)-\left(ii\right)$
$\Rightarrow \frac{C_1}{2}+\frac{C_3}{4}+\cdots +\frac{C_{2019}}{2020}=\frac{2^{2020}-1}{2021}$

Alternate solution:

Let $S_1=C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\cdots +\frac{C_{2020}}{2021}$
$=\sum _0^{2020}(-1{)}^r\frac{{}^{2020}{C}_r}{r+1}$
$=\sum _0^{2020}(-1{)}^r\frac{{}^{2021}{C}_{r+1}}{2021}[\because \frac{{}^{n+1}{C}_{r+1}}{{}^n{C}_r}=\frac{n+1}{r+1}]$
$=\frac{1}{2021}\sum _0^{2020}(-1{)}^r{}^{2021}{C}_{r+1}$
$=-\frac{1}{2021}\sum _0^{2020}(-1{)}^r{}^{2021}{C}_{r+1}(-1{)}^{r+1}[\because \sum _0^n{}^n{C}_r{a}^r=(1+a{)}^n]$
$=-\frac{1}{2021}\left[\right(1-1{)}^{2021}-{}^{2021}{C}_0]=\frac{1}{2021}$

Similarly,

$S_2=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\cdots +\frac{C_{2020}}{2021}$
$=\sum _0^{2020}\frac{{}^{2020}{C}_r}{r+1}$
$=\sum _0^{2020}\frac{{}^{2021}{C}_{r+1}}{2021}$
$=\frac{1}{2021}\sum _0^{2020}{}^{2021}{C}_{r+1}$
$=\frac{1}{2021}[2^{2021}-1][\because \sum _0^n{}^n{C}_r=2^n]$

$S_1+S_2\Rightarrow C_0+\frac{C_2}{3}+\frac{C_4}{5}+\dots +\frac{C_{2020}}{2021}=\frac{2^{2020}}{2021}$

$S_1-S_2\Rightarrow C_1+\frac{C_3}{5}+\frac{C_5}{7}+\dots +\frac{C_{1999}}{2020}=\frac{2^{2020}-1}{2021}$