Question

If \[{C_1}\], \[{C_2}\], \[{C_3}\] are the usual binomial coefficient and \[S = {C_1} + 2{C_2} + 3{C_3} + \cdots + n{C_n}\], then what is the value of \[C\]?
A. \[n{2^n}\]
B. \[{2^{n - 1}}\]
C. \[n{2^{n - 1}}\]
D. \[{2^{n + 1}}\]

Answer 1

Hint: First we will apply the binomial expansion formula of \[{\left( {1 + x} \right)^n} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \cdots + {C_n}{x^n}\]. Then differentiate both sides of the expansion. Then we will put \[x = 1\] in the expansion to get the value of \[S\].

Formula used:
1. Binomial expansion: \[{\left( {1 + x} \right)^n} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \cdots + {C_n}{x^n}\].
2. \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]

Complete step by step solution:
The binomial expansion for \[{\left( {1 + x} \right)^n}\] is
\[{\left( {1 + x} \right)^n} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \cdots + {C_n}{x^n}\]
Differentiate both sides with respect to \[x\].
\[\dfrac{d}{{dx}}{\left( {1 + x} \right)^n} = \dfrac{d}{{dx}}\left( {1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \cdots + {C_n}{x^n}} \right)\]
Now applying the power rule of differentiate
\[n{\left( {1 + x} \right)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} + \cdots + n{C_n}{x^{n - 1}}\]
Now putting \[x = 1\] in the above equation
\[n{\left( {1 + 1} \right)^{n - 1}} = {C_1} + 2{C_2} \cdot 1 + 3{C_3} \cdot {1^2} + \cdots + n{C_n} \cdot {1^{n - 1}}\]
Now simplify the above equation
\[n\,{2^{n - 1}} = {C_1} + 2{C_2} + 3{C_3} + \cdots + n{C_n}\]
Given that \[S = {C_1} + 2{C_2} + 3{C_3} + \cdots + n{C_n}\]
So, \[S = n\,{2^{n - 1}}\]
Hence option C is the correct option.

Note: Many students do a common mistake to calculate the value of \[S\]. They put \[x = 1\] in the expansion \[{\left( {1 + x} \right)^n} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \cdots + {C_n}{x^n}\] which is incorrect way to solve. First, we have to differentiate the expansion then put \[x = 1\] to calculate the value of \[S\].