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

The coefficient of xr[0rn1] in the expression of (x+2)n1+(x+2)n2.(x+1)+(x+2)n3.(x+1)2+...+(x+1)n1 is

A
nCr(2r1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
nCr(2nr1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
nCr(2r+1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
nCr(2nr+1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A nCr(2nr1)
Given expression is equal to np=1(x+3)(np)×(x+2)(p1)
Which is equal to (x+3)(n1)×(np=1((x+2)(x+3))(p1))=(x+3)n(x+2)n
So,coefficient of xr is nCr×(3(nr)2(nr))

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