wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find C0+2C1+3C2+4C3+...........+(n+1)Cn

Open in App
Solution

C0+2C1+3C2+4C3+...........+(n+1)Cn
So, we know that
(1+x)2=C0+C1x+C2x2+C3x3+C4x4+.......+Cnxn(1)
n.(1+n)n1=0+C1+2C2x+3C3x2+.........+nCnx(n1)
diff. with respect to x
n.(1+n)n1=0+C1+2C2x+3C3x2+.........+nCnx(n1)
put x=1 in eq(1) and eq(2)
2n+2(2n1)=C0+2C1+3C2+4C3+.......+(n+1)Cn
2n+n.2n1 Required value

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