Question

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

Answer 1

Hint: Here we will first 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 desired answer.
The general binomial expansion of \[{\left( {a + b} \right)^n}\]is given by:-
\[{\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^n}{\left( b \right)^0}{ + ^n}{C_1}{\left( a \right)^{n - 1}}{\left( b \right)^1}{ + ^n}{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + .............{ + ^n}{C_n}{\left( a \right)^0}{\left( b \right)^n}\]

Complete step-by-step answer:
Here we are given:-
\[{\left( {1 + x} \right)^{23}}\]
Now we know that the general binomial expansion of \[{\left( {a + b} \right)^n}\]is given by:-
\[{\left( {a + b} \right)^n}{ = ^n}{C_0}{\left( a \right)^n}{\left( b \right)^0}{ + ^n}{C_1}{\left( a \right)^{n - 1}}{\left( b \right)^1}{ + ^n}{C_2}{\left( a \right)^{n - 2}}{\left( b \right)^2} + .............{ + ^n}{C_n}{\left( a \right)^0}{\left( b \right)^n}\]
Hence the binomial expansion of \[{\left( {1 + x} \right)^{23}}\]is given by:-
\[{\left( {1 + x} \right)^{23}}{ = ^{23}}{C_0}{\left( 1 \right)^{23}}{\left( x \right)^0}{ + ^{23}}{C_1}{\left( 1 \right)^{23 - 1}}{\left( x \right)^1}{ + ^{23}}{C_2}{\left( 1 \right)^{23 - 2}}{\left( x \right)^2} + .............{ + ^{23}}{C_{23}}{\left( 1 \right)^0}{\left( x \right)^{23}}\]
Solving it further we get:-
\[{\left( {1 + x} \right)^{23}}{ = ^{23}}{C_0}{ + ^{23}}{C_1}{\left( x \right)^1}{ + ^{23}}{C_2}{\left( x \right)^2} + .............{ + ^{23}}{C_{23}}{\left( x \right)^{23}}\]
Now putting \[x = 1\] we get:-
\[{\left( {1 + 1} \right)^{23}}{ = ^{23}}{C_0}{ + ^{23}}{C_1}{\left( 1 \right)^1}{ + ^{23}}{C_2}{\left( 1 \right)^2} + .............{ + ^{23}}{C_{23}}{\left( 1 \right)^{23}}\]
Solving it further we get:-
\[{2^{23}}{ = ^{23}}{C_0}{ + ^{23}}{C_1}{ + ^{23}}{C_2} + .....................{ + ^{23}}{C_{23}}\]
Now we know that:-
\[^{23}{C_0}{ = ^{23}}{C_{23}}\]
\[^{23}{C_1}{ = ^{23}}{C_{22}}\]
\[^{23}{C_2}{ = ^{23}}{C_{21}}\]
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\[^{23}{C_{11}}{ = ^{23}}{C_{12}}\]
Hence, substituting these values we get:-
\[{2^{23}} = 2\left[ {^{23}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]\]
Dividing the equation by 2 we get:-
\[\dfrac{{{2^{23}}}}{2}{ = ^{23}}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}\]
\[ \Rightarrow {2^{22}}{ = ^{23}}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}\]
Now since \[^{23}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}\] is the sum f the coefficients of last 12 terms of the binomial expansion.

Hence, the sum is equal to \[{2^{22}}\].

Note: Students should note that the main trick to solve this question is to put \[x = 1\] in the expansion of \[{\left( {1 + x} \right)^{23}}\] to get the desired answer otherwise it will be difficult to solve the given question.