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

Find the coefficient of x50 in the expression (1+x)1000+2x(1+x)999+3x2(1+x)998++1001x1000

Open in App
Solution

Coefficient of x50 in
S=(1+x)1000+2x(1+x)999+3x2(1+x)998+....+10001x1000
Multiply by ′′x′′1+x
xS(1+x)=x(1+x)999+2x2(1+x)998+.....+1001x10011+x
s[1x1+x]=(1+x)1000+[x(1+x)999+x2(1+x)998+.....+x1000]1001x10011+x
Now
s=x(1+x)999+x2(1+x)998+....+x1000
Multiply by x1+x
S1+x=x2(1+x)998+.....+x10011+x
S1+x=x(1+x)998x10011+x
S=x(1+x)1000x1001
s1+x=(1+x)1000+x(1+x)1000x10011001x10011+x
s=(1+x)1002x1001(1+x)1001x1001
Coefficient of x50 in

(1+x)1002x1001(1+x)1001x1001
=1001C50.

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