Question
Mathematics Question on Binomial theorem
If (1+ax)n=1+6x+227x2+⋯+anxn, then the values of a and n are respectively
A
2,3
B
3,2
C
23,4
D
1,6
Answer
23,4
Explanation
Solution
Given,
(1+ax)n=1+6x+227x2+…+anxn…(i)
The expansion of (1+ax)n is,
(1+ax)n=1+nax+2!n(n−1)(ax)2+…(ii)
On comparing the coefficient of like powers of x in Eqs. (i) and (ii),
na=6…(iii)
227=2n(n−1)⋅a2
⇒27=(n−1)(na)⋅a
27−(n−1)a6 [from E (iii)]
(n−1)a=29…(iv)
From Eqs. (iii) and (iv),
n(n−1)6=29
⇒nn−1=43
⇒4n−4=3n
⇒n=4
From E (iii), a=46
⇒a=3/2