Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```

In the expansion of  , the term which does not contain x is :

1.
10C0

2.
10C7

3.
10C4

4.
none

```
7 years ago

147 Points
```										Dear Vishwesh
simplyfy the given expression
it will become
{(x1/3 +1) -(x-1 + x-1/2)}10
={x1/3 +1 -x-1 -  x-1/2}10
use multinominal theorem and find general term of series
=10!/p!q!r!s! [(x1/3)p(1)q( -x-1 )r(-  x-1/2 )s]
=10!/p!q!r!s! [(xp/3 -r-s/2](-1)r+s   where p+q+r+s =10
for the term does not contain x
p/3 -r-s/2 =0
now find out all possible value
put p=r=s=0
so q=10
so general term become
10C10  or 10C0
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 1,272 off
USE CODE: Neerajwin
Get extra Rs. 187 off
USE CODE: Neerajwin