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

If number of terms in the expansion of (1+4x+x2)n,nN is 1225, then the value of n is-

A
48
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
408
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
612
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1244
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 48
We know that number of terms in expansion of (x1+x2+x3+.....+xr)n is n+r1Cn.
Hence number of terms in (1+4x+x2)n is n+31Cn=n+2C2=1225 (given)
(n+2)(n+1)2=1225(n+2)(n+1)=2450
n2+3n2448=0n=48

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