CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For r=0,1,2,....,n, prove that C0Cr+C1Cr+1+C2Cr+2+....+CnrCn= 2nC(n+r) and hence deduce that
i) C20+C21+C22+......+C2n= 2nCn
ii) C0C1+C1C2+C2C3+.....+Cn1Cn= 2nCn+1

Open in App
Solution

To prove:
C0.Cr+C1.Cr+1+C2.Cr+2+......+Cnr.Cn=2nCn+r
i)C20+C21+C22+........+C2n=2nCn
ii)C0.C1+C1.C2+C2.C3+......+Cn1.Cn=2nCn+1
Solution:
We know that,
C0+C1x+C2x2+......+Cnxn=(1+x)n ...............(1)
C0xn+C1xn1+C2xn2+......+Cn=(x+1)n ...............(2)
Multiplying eqn.(1) and eqn.(2) we get,
(C0+C1x+C2x2+......+Cnxn) (C0xn+C1xn1+C2xn2+......+Cn)=(1+x)2n .............(3)
Equating coeffiecients of xnr from both sides of (3) we get,
C0.Cr+C1.Cr+1+C2.Cr+2+......+Cnr.Cn=2nCn+r ..........(4)
Now putting r=0 in eqn.(4) we get,
C0.C0+C1.C1+C2.C2+......+Cn.Cn=2nCn
or, C20+C21+C22+........+C2n=2nCn
Now putting r=1 in eqn.(4) we get,
C0.C1+C1.C2+C2.C3+......+Cn1.Cn=2nCn+1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Combinations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon