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.