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Question

Mathematics Question on Binomial theorem

If (1+ax)n=1+6x+272x2++anxn\left(1+ax\right)^{n} =1+6x+\frac{27}{2}x^{2}+\cdots+a^{n}\, x^{n}, then the values of aa and nn are respectively

A

2,32, 3

B

3,23, 2

C

32,4\frac{3}{2}, 4

D

1,61, 6

Answer

32,4\frac{3}{2}, 4

Explanation

Solution

Given,
(1+ax)n=1+6x+272x2++anxn(i)(1+a x)^{n}=1+6 x+\frac{27}{2} x^{2}+\ldots+a^{n} x^{n} \,\,\,\,\,\dots(i)
The expansion of (1+ax)n(1+a x)^{n} is,
(1+ax)n=1+nax+n(n1)2!(ax)2+(ii)(1+a x)^{n}=1+n a x+\frac{n(n-1)}{2 !}(a x)^{2}+\ldots (ii)
On comparing the coefficient of like powers of xx in Eqs. (i) and (ii),
na=6(iii)na=6\,\,\,\,\, \dots(iii)
272=n(n1)2a2\frac{27}{2}=\frac{n(n-1)}{2} \cdot a^{2}
27=(n1)(na)a\Rightarrow \, 27 =(n-1)(n a) \cdot a
27(n1)a627 -(n-1) a 6 [from E (iii)]
(n1)a=92(iv)(n-1) a=\frac{9}{2}\,\,\,\,\,\,\dots(iv)
From Eqs. (iii) and (iv),
(n1)6n=92\frac{(n-1) 6}{n}=\frac{9}{2}
n1n=34\Rightarrow\,\frac{n-1}{n}=\frac{3}{4}
4n4=3n\Rightarrow \, 4 n-4=3 n
n=4\Rightarrow \, n=4
From E (iii), a=64a=\frac{6}{4}
a=3/2\Rightarrow \, a=3 / 2