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Question: For a positive integer n, if the mean of the binomial coefficients in the expansion of \[{(a + b)^{2...

For a positive integer n, if the mean of the binomial coefficients in the expansion of (a+b)2n−3  {(a + b)^{2n - 3}}\;is 16, then n is equal to:
A. 5
B. 7
C. 9
D. 4

Explanation

Solution

The binomial expansion is defined as (a+b)n =nC0an +nC1an−1b1 +nC2an−2b2 + ... ... +nCran−rbr+ ... +nCnbn {\left( {a + b} \right)^n}{\text{ }} = {}^n{C_0}{a^n}{\text{ }} + {}^n{C_1}{a^{n-1}}{b^1}{\text{ }} + {}^n{C_2}{a^{n - 2}}{b_2}{\text{ }} + {\text{ }}...{\text{ }}...{\text{ }} + {}^n{C_r}{a^{n-r}}{b_r} + {\text{ }}...{\text{ }} + {}^n{C_n}{b_n}{\text{ }}. To find the mean of coefficients take values of a and b to be equal to 1. The number of terms in the expansion is more than power binomial expression is raised to.

Complete step-by-step answer:
We are given that the mean of the coefficients of binomial expansion is 16. i.e. 2n−3C0+2n−3C1+2n−3C2+.......2n−3Cr.........+2n−3C2n−32n−2=16\dfrac{{{}^{2n - 3}{C_0} + {}^{2n - 3}{C_1} + {}^{2n - 3}{C_2} + .......{}^{2n - 3}{C_r}......... + {}^{2n - 3}{C_{2n - 3}}}}{{2n - 2}} = 16………. (1)
First we need to look at the definition of binomial expression. The algebraic expression which contains only two terms is called binomial. It is a two-term polynomial. Also, it is called a sum or difference between two or more monomials. It is the simplest form of a polynomial.
When this binomial is raised to power of a natural number and expanded, the expansion is called binomial expansion.

The formula for binomial expansion of the question will be –
(a+b)2n−3 =2n−3C0a2n−3 +2n−3C1a2n−4b1 +2n−3C2a2n−5b2 + ... ... +2n−3Cra2n−3−rbr+ ... +2n−3C2n−3b2n−3 {\left( {a + b} \right)^{2n - 3}}{\text{ }} = {}^{2n - 3}{C_0}a{}^{2n - 3}{\text{ }} + {}^{2n - 3}{C_1}a{}^{2n - 4}{b^1}{\text{ }} + {}^{2n - 3}{C_2}{a^{{}^{2n - 5}}}{b^{_2}}{\text{ }} + {\text{ }}...{\text{ }}...{\text{ }} + {}^{2n - 3}{C_r}{a^{2n - 3-r}}{b^{_r}} + {\text{ }}...{\text{ }} + {}^{{}^{2n - 3}}{C_{2n - 3}}{b^{_{{}^{2n - 3}}}}{\text{ }}….. (2)
Whenever the questions involve submission of coefficients of expansion, the best method to solve is taking the terms a and b equal to 1. This is done to eliminate a and b from the expansion.
By taking values of a and b as 1 we only get submission or addition of the coefficients. As shown –
(1+1)2n−3 =2n−3C012n−3 +2n−3C112n−411 +2n−3C212n−512 + ... ... +2n−3Cr12n−3−r1r+ ... +2n−3C2n−312n−3 {\left( {1 + 1} \right)^{2n - 3}}{\text{ }} = {}^{2n - 3}{C_0}1{}^{2n - 3}{\text{ }} + {}^{2n - 3}{C_1}1{}^{2n - 4}{1^1}{\text{ }} + {}^{2n - 3}{C_2}{1^{{}^{2n - 5}}}{1^{_2}}{\text{ }} + {\text{ }}...{\text{ }}...{\text{ }} + {}^{2n - 3}{C_r}{1^{2n - 3-r}}{1^{_r}} + {\text{ }}...{\text{ }} + {}^{{}^{2n - 3}}{C_{2n - 3}}{1^{_{{}^{2n - 3}}}}{\text{ }}
(2)2n−3 =2n−3C0 +2n−3C1 +2n−3C2 + ... ... +2n−3Cr+ ... +2n−3C2n−3 {\left( 2 \right)^{2n - 3}}{\text{ }} = {}^{2n - 3}{C_0}{\text{ }} + {}^{2n - 3}{C_1}{\text{ }} + {}^{2n - 3}{C_2}{\text{ }} + {\text{ }}...{\text{ }}...{\text{ }} + {}^{2n - 3}{C_r} + {\text{ }}...{\text{ }} + {}^{{}^{2n - 3}}{C_{2n - 3}}{\text{ }}
Now, to take out the mean of coefficients we need to divide the equation by the number of terms. Number of terms in an expansion is one more than the power it is raised to.
Therefore, we have to divide 2n−2 2n - 2{\text{ }}i.e. [(2n−3)+1]\left[ {\left( {2n - 3} \right) + 1} \right].
Dividing both side by 2n−2 2n - 2{\text{ }}we get,
(2)2n−32n−2 =2n−3C0 +2n−3C1 +2n−3C2 + ... ... +2n−3Cr+ ... +2n−3C2n−3 2n−2\dfrac{{{{\left( 2 \right)}^{2n - 3}}}}{{2n - 2}}{\text{ }} = \dfrac{{{}^{2n - 3}{C_0}{\text{ }} + {}^{2n - 3}{C_1}{\text{ }} + {}^{2n - 3}{C_2}{\text{ }} + {\text{ }}...{\text{ }}...{\text{ }} + {}^{2n - 3}{C_r} + {\text{ }}...{\text{ }} + {}^{{}^{2n - 3}}{C_{2n - 3}}{\text{ }}}}{{2n - 2}}…… (3)
Now, from equation (1) and equation (3) we have,

\Rightarrow \dfrac{{{{\left( 2 \right)}^{2n - 3}}}}{{2n - 2}} = 16 \\\ \Rightarrow \dfrac{{{{\left( 2 \right)}^{2n}}}}{{n - 1}} = 16 \times 16 \\\ \Rightarrow n = 5 \\\ \end{gathered} $$ Therefore, option (a) 5 is correct. **Note:** There can questions where instead of addition of the coefficients there will be alternate negative signs as shown$${\left( 2 \right)^{2n - 3}}{\text{ }} = {}^{2n - 3}{C_0}{\text{ - }}{}^{2n - 3}{C_1}{\text{ }} + {}^{2n - 3}{C_2}{\text{ - }}{}^{2n - 3}{C_3} + ...{\text{ }}...{\text{ }}...{\text{ }}$$. As such we have to take the value of a as 1 and b as -1.