Binomialized !!

We are given ${ (1+bx) }^{ n }$ in this form : ${ (1+bx) }^{ n }=1+{ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }{ x }^{ 3 }+{ a }_{ 4 }{ x }^{ 4 }+...+(bx)^{ n }\quad \rightarrow \quad (1)$

but we can get another form by using the $binomial$ $theorem$ : ${ (1+bx) }^{ n }=1+{ _{ n }{ C }_{ 1 } }(b{ x) }^{ 1 }+{ _{ n }{ C }_{ 2 } }(b{ x) }^{ 2 }+{ _{ n }{ C }_{ 3 } }({ bx) }^{ 3 }+{ _{ n }{ C }_{ 4 } }(b{ x) }^{ 4 }+...+(bx)^{ n }\quad \rightarrow \quad (2)$ {As "C" refers to the Combination}

From (1),(2) : $1+{ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }{ x }^{ 3 }+{ a }_{ 4 }{ x }^{ 4 }+...+(bx)^{ n }=1+{ _{ n }{ C }_{ 1 } }(b{ x) }^{ 1 }+{ _{ n }{ C }_{ 2 } }(b{ x) }^{ 2 }+{ _{ n }{ C }_{ 3 } }({ bx) }^{ 3 }+{ _{ n }{ C }_{ 4 } }(b{ x) }^{ 4 }+...+(bx)^{ n }$

By equalizing the binomial coefficients : ${ a }_{ 1 }={ _{ n }{ C }_{ 1 } }(b)\quad ,\quad { a }_{ 2 }={ _{ n }{ C }_{ 2 } }{ (b) }^{ 2 }\quad ,\quad { a }_{ 3 }={ _{ n }{ C }_{ 3 } }{ (b) }^{ 3 }\quad ,\quad { a }_{ 4 }={ _{ n }{ C }_{ 4 } }{ (b) }^{ 4 }$ $\boxed { \therefore { a }_{ 1 }=nb=12 } \longmapsto \quad (3)$ Now this one is $important$ , but we'll keep it for now

Well , things got a little bit stuffed right here , so lets clean it up using the information we didn't use yet !

$\because { a }_{ 4 }=4{ a }_{ 2 }\\ \therefore 4{ _{ n }{ C }_{ 2 } }{ (b) }^{ 2 }={ _{ n }{ C }_{ 4 } }{ (b) }^{ 4 }$

$\\ \frac { 4{ (b) }^{ 2 } }{ { (b) }^{ 4 } } =\frac { _{ n }{ C }_{ 4 } }{ _{ n }{ C }_{ 2 } } \\ \\ \frac { 4 }{ { (b) }^{ 2 } } =\frac { _{ n }{ C }_{ 4 } }{ _{ n }{ C }_{ 3 } } \times \frac { _{ n }{ C }_{ 3 } }{ _{ n }{ C }_{ 2 } } \\ \\ \frac { 4 }{ { (b) }^{ 2 } } =\frac { n-4+1 }{ 4 } \times \frac { n-3+1 }{ 3 } \\ \\ \frac { 4 }{ { (b) }^{ 2 } } =\frac { (n-3)(n-2) }{ 12 } \\ \\$

Now , using (3) , we'll substitute $b$ by $\frac { 12 }{ n }$ :

$\frac { 4 }{ { (\frac { 12 }{ n } ) }^{ 2 } } =\frac { { n }^{ 2 }-5n+6 }{ 12 } \\ \\ \frac { { n }^{ 2 } }{ 3 } ={ n }^{ 2 }-5n+6$

After simplifying it should be like this : $2{ n }^{ 2 }-15n+18=0$ , $(2n-3)(n-6)=0$ $\therefore \quad n=\cfrac { 3 }{ 2 } \quad Or\quad 6$ & of course the solution $n=\cfrac { 3 }{ 2 }$ is rejected as n is a positive integer

From (3) we'll find out that $b=2$

$\boxed { \therefore n+b=8 }$