Question

The A.M. of the coefficient in the expression of ${(1 + x)^{30}}$ is: (if n=30)
A) $\dfrac{{{2^{30}}}}{{30}}$
B) $\dfrac{{{2^{30}}}}{{31}}$
C) $\dfrac{{{2^{29}}}}{{11}}$
D) None of these

Answer 1

Hint: Here in this question we have to find arithmetic mean for which we will expand the given expression with the help of binomial expansion then we will find the arithmetic mean which is given by total number of terms divided by number of terms: -
Binomial expansion is given by:-
${(1 + x)^n} = \sum\limits_{r = 0}^n {^n{C_r}.{X^r} = [{C_0} + {C_1}X + {C_2}{X^2} + ...{C_n}{X^n}]} $

Complete step-by-step solution:
First of fall we will put n=30 in the formula of binomial expansion so that we can get the term same as in the question.
$ \Rightarrow {(1 + x)^{30}} = \sum\limits_{r = 0}^{30} {^{30}{C_r}.{X^r} = [{C_0} + {C_1}X + {C_2}{X^2} + ...{C_{30}}{X^{30}}]} $
$ \Rightarrow {(1 + x)^{30}} = [{C_0} + {C_1}X + {C_2}{X^2} + ...{C_{30}}{X^{30}}]$
Now putting X=1 we will get one another equation,
$ \Rightarrow {(2)^{30}} = [{C_0} + {C_1} + {C_2} + ...{C_{30}}]$ .................... Equation 1.
Now numbers of terms which are coefficients of various term in expansion of ${(1 + x)^{30}}$ are 31
Arithmetic mean= Total sum of coefficients/Number of terms
Therefore required mean= $\dfrac{{[{C_0} + {C_1} + {C_2} + ...{C_{30}}]}}{{31}}$
$ \Rightarrow \dfrac{{{{(2)}^{30}}}}{{31}}$ (Putting value from equation 1.)
Therefore A.M. of the coefficient in the expression of ${(1 + x)^{30}}$ is $\dfrac{{{{(2)}^{30}}}}{{31}}$ , so correct option is (B)

Note: Students may likely to do mistake in binomial expression where inside combination formula have been used so below a brief explanation of combination is mentioned:-
Combination: -Number of combination of ‘n’ things has taken ‘r’ at a time then combination formula is given by: -
$C(n,r) = {}^n{C_r}$
$C(n,r) = \dfrac{{n!}}{{r!(n - r)!}}$
There are also two types of combinations (remember the order does not matter):
Repetition is Allowed: such as coins in your pocket (5, 5, 5, 10, 10) so if this 5,5,5,10,10 is selected then it can be selected three times because repetition is allowed.
No Repetition: For example if you are having three balls 1, 2 and 3. They are chosen so possibility is only one 123 while in permutation 6 arrangements can be made For example 123, 321 ,132 ,213, 231 ,312