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Question

If C0,C1,C2,,Cn denote the binomial coefficients respectively in (1+x)2020, then


Correct Answer
A
C0C12+C23C34++C20202021=12021
Your Answer
B
C0+C12+C23+C34++C20202021=2202012021
Correct Answer
C
C0+C23+C45++C20202021=220202021
Your Answer
D
C12+C34++C19992020=220202021

Solution

The correct options are
A C0C12+C23C34++C20202021=12021
C C0+C23+C45++C20202021=220202021
Here Cr= 2020Cr and n=2020

(1+x)2020=C0+C1x+C2x2+C3x3++C2020x2020

Integrating from 0 to 1 both sides,
10(1+x)2020dx=10(C0+C1x+C2x2+C3x3++C2020x2020)dx

[(1+x)20212021]10=[C0x+C1x22+C2x33++C2020x20212021]10

22021202112021=C0+C12+C23++C20202021

C0+C12+C23++C20202021=2202112021   (i)

Similarly,
01(1+x)2020dx=01(C0+C1x+C2x2+C3x3++C2020x2020)dx
[(1+x)20212021]01=[C0x+C1x22+C2x33++C2020x20212021]01
120210=0(C0+C12C23+C20202021)
C0C12+C23C34++C20202021=12021   (ii)

(i)+(ii)
2(C0+C23+C45++C20202021)=220212021
C0+C23+C45++C20202021=220202021

Similarly (i)(ii)
C12+C34++C20192020=2202012021

Alternate solution:

Let S1=C0C12+C23C34++C20202021
=20200(1)r 2020Crr+1
=20200(1)r 2021Cr+12021   [ n+1Cr+1 nCr=n+1r+1]
=1202120200(1)r 2021Cr+1
=1202120200(1)r 2021Cr+1(1)r+1   [n0 nCrar=(1+a)n]
=12021[(11)2021 2021C0]=12021

Similarly, 

S2=C0+C12+C23+C34++C20202021
=20200 2020Crr+1
=20200 2021Cr+12021 
=1202120200 2021Cr+1
=12021[220211]   [n0 nCr=2n]

S1+S2C0+C23+C45++C20202021=220202021

S1S2C1+C35+C57++C19992020=2202012021

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