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

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

Explanation

Solution

Here we will first write the binomial expansion of (1+x)23{\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 (a+b)n{\left( {a + b} \right)^n}is given by:-
(a+b)n=nC0(a)n(b)0+nC1(a)n−1(b)1+nC2(a)n−2(b)2+.............+nCn(a)0(b)n{\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:-
(1+x)23{\left( {1 + x} \right)^{23}}
Now we know that the general binomial expansion of (a+b)n{\left( {a + b} \right)^n}is given by:-
(a+b)n=nC0(a)n(b)0+nC1(a)n−1(b)1+nC2(a)n−2(b)2+.............+nCn(a)0(b)n{\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 (1+x)23{\left( {1 + x} \right)^{23}}is given by:-
(1+x)23=23C0(1)23(x)0+23C1(1)23−1(x)1+23C2(1)23−2(x)2+.............+23C23(1)0(x)23{\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:-
(1+x)23=23C0+23C1(x)1+23C2(x)2+.............+23C23(x)23{\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=1x = 1 we get:-
(1+1)23=23C0+23C1(1)1+23C2(1)2+.............+23C23(1)23{\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:-
223=23C0+23C1+23C2+.....................+23C23{2^{23}}{ = ^{23}}{C_0}{ + ^{23}}{C_1}{ + ^{23}}{C_2} + .....................{ + ^{23}}{C_{23}}
Now we know that:-
23C0=23C23^{23}{C_0}{ = ^{23}}{C_{23}}
23C1=23C22^{23}{C_1}{ = ^{23}}{C_{22}}
23C2=23C21^{23}{C_2}{ = ^{23}}{C_{21}}
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23C11=23C12^{23}{C_{11}}{ = ^{23}}{C_{12}}
Hence, substituting these values we get:-
223=2[23C12+............+23C21+23C22+23C23]{2^{23}} = 2\left[ {^{23}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}} \right]
Dividing the equation by 2 we get:-
2232=23C12+............+23C21+23C22+23C23\dfrac{{{2^{23}}}}{2}{ = ^{23}}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}
⇒222=23C12+............+23C21+23C22+23C23\Rightarrow {2^{22}}{ = ^{23}}{C_{12}} + ............{ + ^{23}}{C_{21}}{ + ^{23}}{C_{22}}{ + ^{23}}{C_{23}}
Now since 23C12+............+23C21+23C22+23C23^{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 222{2^{22}}.

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