Question

the first three terms in the expansion of (1+ax)^n are 1+12 x+64 x^2.find n and a.

Answer 1

remember binomial theorem?

Answer 2

check it please how can you make from 1+12x +64x^2 one perfect square using formula
(a+b)^2 = a^2 +2ab +b^2

Answer 3

\[\left( 1+ax \right)^n=nc0*1^n+nc1*1^{n-1}\left( ax \right)^1+nc2*1^{n-2}\left( ax \right)^2+...\]
\[=1+nax+\frac{ n \left( n-1 \right) }{ 2*1}a^2x^2+...\]
nax=12x
\[\frac{ n \left( n-1 \right) a^2 x^2}{ 2 }=64x^2\]
na=12 ...(1)
again
\[\frac{ n \left( n-1 \right)a^2x^2 }{ 2n^2a^2x^2 }=\frac{ 64x^2 }{ 144 x^2}\]
\[\frac{ \left( n-1 \right) }{ 2n }=\frac{ 64 }{ 144 }=\frac{ 4 }{ 9 }\]
9n-9=8n
n=9
na=12
9a=12
\[a=\frac{ 12 }{ 9 }=\frac{ 4 }{ 3 }\]