CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
156
You visited us 156 times! Enjoying our articles? Unlock Full Access!
Question

If ak is the coefficient of xk in the expansion of (1+x+x2)n for k=0,1,2, , 2n, then a1+2a2+3a3++2na2n=

A
a0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
n.3n
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
n3n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C n.3n
(1+x+x2)n=1+a1x+a2x2+a3x3...+a2nx2n
Differentiating the above expression with respect to x, we get
n(1+x+x2)n1(2x+1)=a1+2a2x+3a3x2+4a4x3...+2na2nx2n1
Let x=1,
Therefore, we get
n3n1(3)=a1+2a2+3a3...+2na2n
a1+2a2+3a3..+2na2n=n.3n

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