Question

If (1+ax)n=1+8x+24x2......., then the values of a and n are equal to

Answer 1

Please see the attachment

Answer 2

Answer:

n = 4

and,

a = 2

Step-by-step explanation:

According to the question,

We have the Binomial expansion of the equation,

(1 + ax)ⁿ = 1 + 8x + 24x² + ..................

So,

We know that the Binomial expansion of the term of the form, (1 + ax)ⁿ, is given by,

$(1 + ax)^{n}=1+^{n}C_{1}(ax)+^{n}C_{2}(ax)^{2}+...........\\(1 + ax)^{n}=1+8x+24x^{2}+.............$

Therefore, on comparing both the equations we can say that,

$^{n}C_{1}(ax)=8x\\So,\\^{n}C_{1}a=8\\\frac{n!}{(n-1)!}a=8\\na=8 ............(1)$

Also, from the 2nd term we can say that,

$^{n}C_{2}(ax)^{2}=24x^{2}\\^{n}C_{2}(a)^{2}=24\\\frac{n!}{2!(n-2)!}(a)^{2}=24\\n(n-1)a^{2}=48..........(2)$

Therefore, from both the eqns. (1) and (2), we can say that,

$n(n-1)a^{2}=48\\n^{2}a^{2}-na^{2}=48\\na(na-a)=48$

On putting the value of 'na' we get,

$na(na-a)=48\\8(8-a)=48\\8-a=6\\a=2$

and,

n = 4

Therefore, the value of 'n' and 'a' is 4 and 2 respectively.