Question

In the binomial expansion of 3√2+1/3^1/3
the ratio of the 7th term from the begining to the 7th term
from the end is 1:6; find n.​

Answer 1

Answer:

n = 6

Step-by-step explanation:

as we know,

the term which we want to find out from the beginning and from the end respectively is as follows

T7th term from the beginning = nC6 [(2)^1/3]^n-6* [(3)-1/3]^6

T7th term from the end = nCn-6 [(2)^1/3]^6* [(3)^-1/3]^n-6

so 1/6 = {nC6 [(2)^1/3]^n-6* [(3)-1/3]^6}/{nCn-6 [(2)^1/3]^6* [(3)^-1/3]^n-6}

so we cancelled out the term nC6 and nCn-6 from the numerator and denominator. and now we are left with

1/6 = 2^[((n-6)/3)-6/3]*3^[(-6/3+(n-6)/3)]

1/6 = 6^[((n-6)/3)-6/3]

6^(-1) = (6)^[((n-6)/3)-6/3]

so by comparing the power, we get

[((n-6)/3)-6/3] = -1

(n-6/3)-2 = -1

by solving this ,we get n = 9

Answer 2

Answer is in the attachment