Question

The value of $ a{C_0} + (a + b){C_1} + (a + 2b){C_2} + ....... + (a + nb){C_n} $ Is equal to
A. $ (2a + nb){2^n} $
B. $ (2a + nb){2^{n - 1}} $
C. $ (na + 2b){2^n} $
D. $ (na + 2b){2^{n - 1}} $

Answer 1

Hint: In these types of questions the students are advised to understand the combination theory.
This type of questions involve the series combination involving the coefficients ranging till the n terms.
Students are advised to go through the series expansion theory to attempt such type of questions.
This is an example of a binomial theorem.
Binomial theorem used in this example because it is used to expand the equation with powers.
 $ {(1 = x)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  k
\end{array}} \right)} {x^k} $
According to the theorem, it is possible to expand the polynomial $ {(x + y)^n} $ into a sum involving terms of the form $ a{x^b}{y^c} $ ,
where the exponents b and c are nonnegative integers with $ b + c = n $ , and the coefficients of each term is a specific positiveinteger depending on \[n\,and\,b.\]

Complete step-by-step answer:
From the given question,
We have the equation,
 $ a{C_0} + (a + b){C_1} + (a + 2b){C_2} + ....... + (a + nb){C_n} $
Where,
 $ {C_{0,}}{C_{1,}}........,{C_N} $ denote the binomial coefficients.
Now, let's get back to the equation,
 $ a{C_0} + (a + b){C_1} + (a + 2b){C_2} + ....... + (a + nb){C_n} $
According to Binomial Theorem we know that,
 $ \sum\limits_{r = 0}^n {(a + rb){}^n} {C_r} $ is true for some r
The calculations follow as below
We will expand the above form because
 $ a{C_0} + (a + b){C_1} + (a + 2b){C_2} + ....... + (a + nb){C_n} $ is equal to $ \sum\limits_{r = 0}^n {(a + rb){}^n} {C_r} $
Thus, we get the following derivation by step by step approaching the problem.
 $
  \sum\limits_{r = 0}^n {a{}^n} {C_r} + \sum\limits_{r = 0}^n {rb{}^n} {C_r} \\
   \Rightarrow a(\sum\limits_{r = 0}^n {{}^n} {C_r}) + {b^n}(\sum\limits_{r = 0}^n {r{}^n} {C_r}) \\
   \Rightarrow a(\sum\limits_{r = 0}^n {{}^n} {C_r}) + {b^n}(\sum\limits_{r = 0}^n {r.\dfrac{n}{r}{}^{n - 1}} {C_{r - 1}}) \\
   \Rightarrow a\sum\limits_{r = 0}^n {{}^n} {C_r} + {b^n}\sum\limits_{r = 0}^n {{}^{n - 1}} {C_{r - 1}} \\
   \Rightarrow a{2^n} + bn{2^{n - 1}} \\
   \Rightarrow (2a + nb){2^{n - 1}}
$

So, the correct answer is “Option B”.

Note: Students might make mistakes while dealing with the operations that involve use of summation and combination.
Students will understand the use of the terms \[r{\text{ }}and{\text{ }}C\] after revising the Binomial theorem.
Students must understand what is asked in such a question and should be able to classify the data correctly.
Students might mess up while putting the values in the actual equation.