# Use the Binomial Theorem to find the coefficient of x^7 in the expansion of (3x + 1)^{12}.

## Question:

Use the Binomial Theorem to find the coefficient of {eq}x^7 {/eq} in the expansion of {eq}(3x + 1)^{12}. {/eq}

## Binomial Theorem :

Binomial theorem gives a way for the expansion of {eq}\left( x+y \right)^n {/eq} for some positive number {eq}n {/eq}.

The binomial expansion is given by

{eq}{\left( {x + y} \right)^n}{ = ^n}{C_0}{\left( x \right)^0}{\left( y \right)^n}{ + ^n}{C_1}{\left( x \right)^1}{\left( y \right)^{n - 1}}{ + ^n}{C_2}{\left( x \right)^2}{\left( y \right)^{n - 2}}...................{ + ^n}{C_n}{\left( x \right)^n}{\left( y \right)^0} = \sum\limits_{r = 0}^n {^n{C_r}{{\left( x \right)}^r}{{\left( y \right)}^{n - r}}} {/eq}

We know,

Using the binomial theorem, the expansion of {eq}\left( 1+ax \right)^n {/eq} is given by {eq}{\left( {ax + 1} \right)^n}{ = ^n}{C_0}{\left( {ax} \right)^0}{\left( 1 \right)^n}{ + ^n}{C_1}{\left( {ax} \right)^1}{\left( 1 \right)^{n - 1}}{ + ^n}{C_2}{\left( {ax} \right)^2}{\left( 1 \right)^{n - 2}}...................{ + ^n}{C_n}{\left( {ax} \right)^n}{\left( 1 \right)^0} {/eq}

The general term of this expansion is {eq}^n{C_r}{\left( {ax} \right)^r}{\left( 1 \right)^{n - r}} {/eq}

Therefore, the expansion of {eq}\left( 3x+1 \right)^{12} {/eq} will be {eq}{\left( {3x + 1} \right)^{12}}{ = ^{12}}{C_0}{\left( {3x} \right)^0}{\left( 1 \right)^{12}}{ + ^{12}}{C_1}{\left( {3x} \right)^1}{\left( 1 \right)^{11}}{ + ^{12}}{C_2}{\left( {3x} \right)^2}{\left( 1 \right)^{10}}...................{ + ^{12}}{C_{12}}{\left( {3x} \right)^{12}}{\left( 1 \right)^0} {/eq} and the general term will be {eq}^{12}{C_r}{\left( {3x} \right)^r}{\left( 1 \right)^{12 - r}} {/eq}

Therefore, to find the coefficient of {eq}x^{7} {/eq}, we should have {eq}r=7 {/eq}.

Hence, the term involving {eq}x^{7} {/eq} is {eq}^{12}{C_7}{\left( {3x} \right)^7}{\left( 1 \right)^{12 - 7}} {/eq}

Hence, the coefficient is given by {eq}^{12}{C_7}{\left( 3 \right)^7} {/eq}