Question

Find the value of \[{1^2} \cdot {C_1} + {3^2} \cdot {C_3} + {5^2} \cdot {C_5} + \cdots \]
A. \[n{\left( {n - 1} \right)^{n - 2 + n \cdot {2^{n - 1}}}}\]
B. \[n{\left( {n - 1} \right)^{n - 2}}\]
C. \[n{\left( {n - 1} \right)^{n - 3}}\]
D. None of these

Answer 1

Hint: We will differentiate the binomial expansion of \[{\left( {1 + x} \right)^n}\] two times with respect to x. Then we substitute x= 1 and x =-1 and add both equations to get the required result.

Formula Used:Binomial expansion of \[{\left( {1 + x} \right)^n} = 1 + {C_1} \cdot x + {C_2} \cdot {x^2} + {C_3} \cdot {x^3} + \cdots + {C_n} \cdot {x^n}\]


Complete step by step solution: Binomial expansion of
\[{\left( {1 + x} \right)^n} = 1 + {C_1} \cdot x + {C_2} \cdot {x^2} + {C_3} \cdot {x^3} + \cdots + {C_n} \cdot {x^n}\]
Now differentiate the above equation with respect to x:
\[\dfrac{d}{{dx}}{\left( {1 + x} \right)^n} = \dfrac{d}{{dx}}\left( {1 + {C_1} \cdot x + {C_2} \cdot {x^2} + {C_3} \cdot {x^3} + \cdots + {C_n} \cdot {x^n}} \right)\]
Applying the power rule of differentiation:
\[\dfrac{d}{{dx}}{\left( {1 + x} \right)^n} = \dfrac{d}{{dx}}\left( {1 + {C_1} \cdot x + {C_2} \cdot {x^2} + {C_3} \cdot {x^3} + \cdots + {C_n} \cdot {x^n}} \right)\]
\[ \Rightarrow n{\left( {1 + x} \right)^{n - 1}} = {C_1} + 2{C_2} \cdot x + 3{C_3} \cdot {x^2} + \cdots + n{C_n} \cdot {x^{n - 1}}\]
Multiplying both sides by x:
\[ \Rightarrow n{\left( {1 + x} \right)^{n - 1}}x = {C_1}x + 2{C_2} \cdot {x^2} + 3{C_3} \cdot {x^3} + \cdots + n{C_n} \cdot {x^n}\]
Differentiate the above equation with respect to x:
\[ \Rightarrow n{\left( {1 + x} \right)^{n - 1}} + n\left( {n - 1} \right){\left( {1 + x} \right)^{n - 2}}x = {C_1} + {2^2}{C_2} \cdot x + {3^2}{C_3} \cdot {x^2} + \cdots + {n^2}{C_n} \cdot {x^{n - 1}}\] ….(i)
Now substituting x = 1 in equation (i)
\[ \Rightarrow n{2^{n - 1}} + n\left( {n - 1} \right){2^{n - 2}} = {C_1} + {2^2}{C_2} + {3^2}{C_3} + \cdots + {n^2}{C_n}\] ….(ii)
Now substituting x = -1 in equation (i)
\[ \Rightarrow 0 = {C_1} - {2^2}{C_2} + {3^2}{C_3} - \cdots \] ….. (iii)
Adding equation (ii) and (iii)
\[ \Rightarrow 2\left( {{C_1} + {3^2}{C_3} + {5^2}{C_5} \cdots } \right) = n{2^{n - 1}} + n\left( {n - 1} \right){2^{n - 2}}\]
Dividing both sides by 2
\[ \Rightarrow {C_1} + {3^2}{C_3} + {5^2}{C_5} + \cdots = \dfrac{{n{2^{n - 1}} + n\left( {n - 1} \right){2^{n - 2}}}}{2}\]
\[ \Rightarrow {C_1} + {3^2}{C_3} + {5^2}{C_5} + \cdots = n{2^{n - 2}} + n\left( {n - 1} \right){2^{n - 3}}\]
Taking common \[n \cdot {2^{n - 3}}\] from the right expression of the equation:
\[ \Rightarrow {C_1} + {3^2}{C_3} + {5^2}{C_5} + \cdots = n{2^{n - 3}}\left( {2 + n - 1} \right)\]
\[ \Rightarrow {C_1} + {3^2}{C_3} + {5^2}{C_5} + \cdots = {2^{n - 3}}n\left( {n + 1} \right)\]


Option ‘D’ is correct

Note: To solve the given question we have to multiply x on both sides of the first derivative of the binomial expansion. Students often forgot to multiply x.