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

Number of terms free from radical sign in the expansion of (1+31/3+71/2)10 is

A
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 8
(1+(313+712))10
Let
(313+712)=x
Then the above expression reduces to,
(1+x)10
Now writing the general term, we get
Tr+1=10Crxr
=10Cr(313+712)r.
Hence we get one rational term for each
r=0,2,3,4,6,8,9,10
Hence in total 8 terms.

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