Question

Sum of the last $12$ coefficients in the binomial expansion of ${\left( {1 + x} \right)^{23}}$ is:

Answer 1

Here we will write the binomial expansion of ${\left( {1 + x} \right)^{23}}$ and then we will put $x = 1$ and then solve it further to get the required answer.

Complete step-by-step answer:
It is given that the binomial expansion is ${\left( {1 + x} \right)^{23}}...\left( 1 \right)$
We have to find the sum of the last $12$ coefficients in the given binomial expansion.
Here we have to use the general binomial expansion of ${\left( {a + b} \right)^n}$ is defined by,
${\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^n}{\left( b \right)^0}{ + ^n}{C_1}{\left( a \right)^{n - 1}}{\left( b \right)^1}{ + ^n}{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + .....{ + ^n}{C_n}{\left( a \right)^0}{\left( b \right)^n}$
Now we elaborate for given the binomial expansion of ${\left( {1 + x} \right)^{23}}$ is written as,
${\left( {1 + x} \right)^{23}}{ = ^{23}}{C_0}{\left( 1 \right)^{23}}{\left( x \right)^0}{ + ^{23}}{C_1}{\left( 1 \right)^{23 - 1}}{\left( x \right)^1}{ + ^{23}}{C_2}{\left( 1 \right)^{23 - 2}}{\left( x \right)^2} + .....{ + ^{23}}{C_{23}}{\left( 1 \right)^0}{\left( x \right)^{23}}$
Now we have to subtract on its power term we get,
${\left( {1 + x} \right)^{23}}{ = ^{23}}{C_0}{\left( 1 \right)^{23}}{\left( x \right)^0}{ + ^{23}}{C_1}{\left( 1 \right)^{22}}{\left( x \right)^1}{ + ^{23}}{C_2}{\left( 1 \right)^{21}}{\left( x \right)^2} + .....{ + ^{23}}{C_{23}}{\left( 1 \right)^0}{\left( x \right)^{23}}$
Here the value $1$ has any power is equal to its value that is $1$
${\left( {1 + x} \right)^{23}}{ = ^{23}}{C_0}{\left( x \right)^0}{ + ^{23}}{C_1}{\left( x \right)^1}{ + ^{23}}{C_2}{\left( x \right)^2} + .....{ + ^{23}}{C_{23}}{\left( x \right)^{23}}$
Now we have to put $x = 1$ we get,
${\left( {1 + 1} \right)^{23}}{ = ^{23}}{C_0}{\left( 1 \right)^0}{ + ^{23}}{C_1}{\left( 1 \right)^1}{ + ^{23}}{C_2}{\left( 1 \right)^2} + .....{ + ^{23}}{C_{23}}{\left( 1 \right)^{23}}$
On adding the LHS and the value $1$ have any power is equal to its value that is $1$, we get
${\left( 2 \right)^{23}}{ = ^{23}}{C_0}{ + ^{23}}{C_1}{ + ^{23}}{C_2} + .....{ + ^{23}}{C_{23}}...\left( 2 \right)$
Now we already know that,
$^{23}{C_0}{ = ^{23}}{C_{23}}$
$^{23}{C_1}{ = ^{23}}{C_{22}}$
$^{23}{C_2}{ = ^{23}}{C_{21}}$
.
.
.
$^{23}{C_{11}}{ = ^{23}}{C_{12}}$
Now we substituting these values in equation $\left( 2 \right)$ we get,
${\left( 2 \right)^{23}} = 2\left[ {^{23}{C_{12}} + ...{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]$
Dividing the above equation by $2$ we get,
$\dfrac{{{{\left( 2 \right)}^{23}}}}{2} = \left[ {^{23}{C_{12}} + ...{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]$
Now we take LHS, the denominator term turns into the numerator, we have to subtract it power value we get,
${\left( 2 \right)^{22}} = \left[ {^{23}{C_{12}} + ...{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]$
Now, $\left[ {^{23}{C_{12}} + ...{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]$is the sum of the coefficients of last $12$ terms of the binomial expansion${\left( {1 + x} \right)^{23}}$.
Hence, the sum is equal to${\left( 2 \right)^{22}}$.

$\therefore$The sum of the coefficient of last $12$ terms of the binomial expansion ${\left( {1 + x} \right)^{23}}$ is equal to ${\left( 2 \right)^{22}}$

Note: In the question we have to remember that the progress from the first term to the last, the exponent of a decrease by $1$ form term to term.
While the exponent of $b$ increases by $1$.
In addition, the sum of the exponents of $a$ and $b$ in each term is $n$
If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that terms, we obtain the coefficient of the next term.