IIT JEE , AIEEE Forum - Mathematics , Algebra

sumit_grg

the cofficient of 3 consecutive term in the expansion of (1+x)ⁿ are in the ratio 1:7:35 the

a> n is divisible by 5 b>n is not divisible by any other than itself

c> n is divisible by 35 d> n is divisible by 23

ashish_banga

answer is B
first coefficient of the expansion = (n,0) = 1
so the coefficients of the consecutive terms start from 1
7 is the coefficient of the second term of the expansion
and coefficient of 3 term = 35
(n,1) = 7
(n,2) = 35
solve these two
n =11

sumit_grg

no it is not the right answer

anchitsaini

Let the coefficients be
ⁿC_r , ⁿC_r+1 , ⁿC_r+2

using the property -->

ⁿC_r /ⁿC_r-1 = (n-r+1) / r

ⁿC_{r +1} / ⁿC_r = (n - r) / (r + 1)

                  = 7

hence

n = 8r + 7   ---1

Also ,

ⁿC_{r +2} / ⁿC_r+1 = (n - r - 1) / (r + 2)

                     =35/7 = 5

hence

n = 6r + 11 ----2

from 1 and 2 ,

r = 2 , n = 23

hence option d )

sriram.a

ans is D

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