Question

In the expansion the number of terms free from radicals is

Answer 1

\[\huge (7^{1/3} + 11^{1/9})^{6561}\]

Answer 2

so basically it should be a multiple of 9

Answer 3

i am getting 729 is it correct or not

Answer 4

looks good!

Answer 5

the number of terms should be 1.....

Answer 6

but answer is 730 =(

Answer 7

then we need to think a bit more

Answer 8

notice that the exponent is not same for both the terms

Answer 9

9 is just 3*3

Answer 10

they split as

k, n-k right ?

Answer 11

what

Answer 12

yes yes

Answer 13

\[\binom{n}{k} a^{n-k}b^{k}\]

Answer 14

got what you are saying

Answer 15

you want n-k to be multiple of 3 and k to be multiple of 9

Answer 16

yes

Answer 17

but how do we find k

Answer 18

it changes continiously

Answer 19

k changes @ganeshie8

Answer 20

I have seen some use of integration in this chapter , if that's the case here i am not fully aware of calculus

Answer 21

@ganeshie8 r u there

Answer 22

yes it has got nothing to do with integration

Answer 23

no problem then

Answer 24

but how do you find k

Answer 25

@No.name
Can you tell me the range of k in \(\binom{n}{k} a^{n-k}b^{k}\)?

Answer 26

`you want n-k to be multiple of 3 and k to be multiple of 9`

6561-k = 3a
k = 9b

Answer 27

notice that 6561 is divisible by 3

Answer 28

6561-k = 3a
k = 9b

is same as

k = 3c
k = 9b

yes ?

Answer 29

what is c ?

Answer 30

yes ofc

Answer 31

try this maybe :
would you agree whenever k is divisible by 9, that makes n-k also divisble by 3

Answer 32

yes

Answer 33

because n = 6561 here which is divisible by 3
so if k is divisible by 9, then for sure n-k is also divisible by 3

Answer 34

yes ofcourse

Answer 35

just find how many integers between 0 and 6561 are divisible by 9